1 Introduction

Let G be a finite molecular graph of order n and size m with vertex set \(V(G)\) and edge set \(E(G)\). The degree of a vertex \(x\in V(G)\), denoted by \(d_{G}(x)\), is the number of edges adjacent to x. A vertex with degree 1 is called a pendent vertex. The minimum and maximum degrees of G are respectively defined by \(\bigtriangleup _{G}=\max\{d_{G}(x): x\in V(G)\}\) and \(\delta_{G}=\min\{d_{G}(x):x\in V(G)\}\). The set of neighboring vertices of a vertex x is denoted by \(N_{G}(x)\). The graph obtained by deleting a vertex \(x\in V(G)\) is denoted by \(G-x\). The graph obtained from G by adding an edge xy between two nonadjacent vertices \(x,y\in V(G)\) is denoted by \(G+xy\).

Historically, the first degree based topological indices are the Zagreb indices, introduced by Gutman and Trinajestić [1]. The first Zagreb index \(M_{1}(G)\) and the second Zagreb index \(M_{2}(G)\) are defined as

$$ M_{1}(G)=\sum_{x\in V(G)}\bigl(d_{G}(x) \bigr)^{2}\quad \mbox{and} \quad M_{2}(G)=\sum _{xy\in E(G)}d_{G}(x)d_{G}(y). $$

The general Randić index (or product-connectivity index) was proposed by Bollobás and Erdős [2] and is defined as follows:

$$ R_{\alpha}(G)=\sum_{xy\in E(G)}\bigl(d_{G}(x)d_{G}(y) \bigr)^{\alpha}, $$

where α is a real number. Then \(R_{-1/2}(G)=\sum_{xy\in E(G)}(d_{G}(x)d_{G}(y))^{-\frac{1}{2}}\) is called the Randić index, which was defined by Randić [3], and \(R_{1}\) is the second Zagreb index. The Randić index is the most studied, most applied, and most important index among all topological indices. Recently, Zhou and Trinajstić [4] modified the concept of Randić index and obtained a new index, called the general sum-connectivity index and defined as follows:

$$ \chi_{\alpha}(G)=\sum_{xy\in E(G)} \bigl(d_{G}(x)+d_{G}(y)\bigr)^{\alpha}, $$

where α is a real number. Then \(\chi_{-1/2}(G)= \sum_{xy\in E(G)}(d_{G}(x)+d_{G}(y))^{-\frac{1}{2}}\) is the sum-connectivity index defined by Zhou and Trinajstić [5], and if \(\alpha=1\), then the general sum-connectivity index becomes the first Zagreb index.

A connected graph G is a cactus if each block of G is either a cycle or an edge. Let \(X(n,t)\) be the set of cacti of order n with t cycles. Obviously, \(X(n,0)\) is the set of trees of order n, and \(X(n,1)\) is the set of unicyclic graphs of order n. Denote by \(X^{0}(n,t)\) the n-vertex cactus consisting of t triangles and \(n-2t-1\) pendent edges such that triangles and pendent edges have exactly one vertex in common.

Lin et al. [6] discuss the sharp lower bounds of the Randić index of cacti with r pendent vertices. Dong and Wu [7] calculated the sharp bounds on the atom-bond connectivity index in the set of cacti with t cycles and also in the set of cacti with r pendent vertices, for all positive integral values of t and r. Li [8] determined the unique cactus with maximum atom-bond connectivity index among cacti with n vertices and t cycles, where \(0\leq t\leq\lfloor\frac{n-1}{2}\rfloor\), and also among n vertices and r pendent vertices, where \(0\leq r\leq n-1\). Ma and Deng [9] calculated the sharp lower bounds of the sum-connectivity index in the set of cacti of order n with t cycles and also in the set of cacti of order 2n with perfect matching. Lu et al. [10] gave the sharp lower bound on the Randić index of cacti with t number of cycles.

The present paper is motivated by the results of papers [9, 10]. The goal of this paper is to compute the sharp lower bounds for the general sum-connectivity index and the general Randić index of cacti with fixed number of cycles.

2 General sum-connectivity index

In this section, we find sharp lower bound for the general sum-connectivity index of cacti. Let

$$ F(n,t)=2t(n+1)^{\alpha}+(n-2t-1)n^{\alpha}+4^{\alpha}t, $$

where \(n\geq3\) and \(t\geq0\). First, we give some lemmas that will be used in the main result.

Lemma 2.1

[4]

Let \(n\geq5\) be a positive integer, \(\alpha<0\) be a real number, and \(G\in X(n,0)\). Then

$$ \chi_{\alpha}(G)\geq(n-1)n^{\alpha}. $$

Equality holds if and only if \(G\cong X^{0}(n,0)\).

Lemma 2.2

[11]

Let \(n\geq5\) be a positive integer, \(-1\leq\alpha<0\) be a real number, and \(G\in X(n,1)\). Then

$$ \chi_{\alpha}(G)\geq2(n+1)^{\alpha}+(n-3)n^{\alpha}+4^{\alpha}. $$

Equality holds if and only if \(G\cong X^{0}(n,1)\).

Lemma 2.3

Let \(x,k\) be positive integers with \(x\geq k\geq1\), and \(\alpha<0\) be a real number. Define

$$ f(x)=x^{\alpha}-(x-k)^{\alpha}. $$

Then \(f(x)\) is an increasing function.

Proof

It is easily seen that \(f'(x)=\alpha(x^{\alpha-1}-(x-k)^{\alpha-1})>0\) for \(\alpha<0\). Therefore, \(f(x)\) is an increasing function. □

Lemma 2.4

Let \(x,k\) be positive integers with \(x\geq k\geq1\), and \(-1\leq\alpha <0\) be a real number. Define

$$ f(x)=k(x+1)^{\alpha}+(x-k) (x+2)^{\alpha}-(x-k) (x+2-k)^{\alpha}. $$

Then \(f(x)\) is a decreasing function.

Proof

Let \(g(x)=k(x+1)^{\alpha}-k(x+2)^{\alpha}\) for \(x\geq k\geq1\). We get \(g'(x)={\alpha}k ((x+1)^{\alpha-1}-(x+2)^{\alpha-1} )<0\). So, \(g(x)\) is a decreasing function.

Let \(h(x)=x(x+2)^{\alpha}\) for \(x\geq k\geq1\). Then \(h''(x)=\alpha((\alpha +1)x+4)(x+2)^{\alpha-2}<0\) for \(-1\leq\alpha<0\). Hence, \(h(x)-h(x-k)\) is a decreasing function. Note that \(f(x)=g(x)+(h(x)-h(x-k))\). Thus, \(f(x)\) is a decreasing function. □

Lemma 2.5

Let \(x,k\) be positive integers with \(x\geq k\geq1\), and \(-1\leq\alpha <0\) be a real number. Define

$$ f(x)=x(x+2)^{\alpha}-(x-2)x^{\alpha}. $$

Then \(f(x)\) is a decreasing function.

Proof

Let \(h(x)=x(x+2)^{\alpha}\) for \(x\geq k\geq1\). Then \(h''(x)=\alpha((\alpha +1)x+4)(x+2)^{\alpha-2}<0\) for \(-1\leq\alpha<0\). But \(f(x)=h(x)-h(x-2)\). Thus, \(f(x)\) is a decreasing function. □

Theorem 2.1

Let \(-1\leq\alpha<0\) be a real number, \(n\geq3\) be a positive integer, and \(G\in X(n,t)\). Then

$$ \chi_{\alpha}(G)\geq F(n,t). $$
(2.1)

Equality holds if and only if \(G\cong X^{0}(n,t)\).

Proof

We use mathematical induction on n and t. If \(t=0\) or \(t=1\), then Lemmas 2.1 and 2.2 give inequality (2.1). If \(n=5\) and \(t=2\), then there is only one possibility that \(X^{0}(5,2)\) is a graph with two cycles that have only one common vertex (see Figure 1) and inequality (2.1) holds. If \(n\geq6\) and \(t\geq2\), then we will discuss the following two cases.

Figure 1
figure 1

\(\pmb{X^{0}(5,2)}\) .

Full size image

Case 1: \(G\in X(n,t)\) has at least one pendent vertex.

Let \(x\in V(G)\) be a pendent vertex and adjacent with vertex y of degree d, where \(2\leq d\leq n-1\). The set of neighbors of y in G is \(N_{G}(y)\setminus\{x\}=\{u_{1},u_{2},u_{3},\dots,u_{d-1}\}\). Without lost of generality, we assume that \(u_{1},u_{2},u_{3},\dots,u_{k-1}\) are pendent vertices and \(u_{k},u_{k+1},\dots,u_{d-1}\) are nonpendent vertices, where \(k\geq1\). If \(k=1\), then \(u_{1},u_{2},u_{3},\dots,u_{d-1}\) are nonpendent vertices. Obviously, we have \(t\leq\lfloor\frac{n-k-1}{2}\rfloor\). If H is a graph obtained by deleting vertices \(x,u_{1},u_{2},\dots,u_{k-1}\) from G, that is, \(H=G-x-u_{1}-u_{2}-\cdots-u_{k-1}\), then \(H\in X(n-k,t)\). By inductive assumption and using Lemmas 2.3 and 2.4, we obtain

$$\begin{aligned} \chi_{\alpha}(G) =&\chi_{\alpha}(H)+k(d+1)^{\alpha}+\sum _{i=k}^{d-1} \bigl(\bigl(d+d_{G}(u_{i}) \bigr)^{\alpha}-\bigl(d-k+d_{G}(u_{i}) \bigr)^{\alpha } \bigr) \\ \geq&F(n-k,t)+k(d+1)^{\alpha}+\sum_{i=k}^{d-1} \bigl(\bigl(d+d_{G}(u_{i})\bigr)^{\alpha}- \bigl(d-k+d_{G}(u_{i})\bigr)^{\alpha} \bigr) \\ =&F(n,t)+2t(n-k+1)^{\alpha}-2t(n+1)^{\alpha}+(n-k-2t-1) (n-k)^{\alpha}-(n-2t-1)n^{\alpha}\\ &{}+k(d+1)^{\alpha}+\sum_{i=k}^{d-1} \bigl(\bigl(d+d_{G}(u_{i})\bigr)^{\alpha }- \bigl(d-k+d_{G}(u_{i})\bigr)^{\alpha} \bigr) \\ \geq& F(n,t)+2t(n-k+1)^{\alpha}-2t(n+1)^{\alpha}+(n-k-2t-1) (n-k)^{\alpha}-(n-2t-1)n^{\alpha}\\ &{}+k(d+1)^{\alpha}+(d-k) \bigl((d+2)^{\alpha}-(d-k+2)^{\alpha} \bigr) \\ \geq& F(n,t)+2t(n-k+1)^{\alpha}-2t(n+1)^{\alpha}+(n-k-2t-1) (n-k)^{\alpha}-(n-2t-1)n^{\alpha}\\ &{}+kn^{\alpha}+(n-1-k) (n+1)^{\alpha}-(n-1-k) (n+1-k)^{\alpha} \\ =&F(n,t)+(n-1-k-2t) \bigl((n+1)^{\alpha}-(n-k+1)^{\alpha}-n^{\alpha }+(n-k)^{\alpha} \bigr) \\ \geq&F(n,t). \end{aligned}$$

The equality holds if and only if \(d=n-1\), \(2t=n-k-1\), and \(H\cong X^{0}(n-k,t)\). Therefore, we have that \(\chi_{\alpha}(G)=F(n,t)\) if and only if \(G\cong X^{0}(n,t)\).

Case 2: \(G\in X(n,t)\) has no pendent vertex. Then we consider edges \(x_{0}x_{1}, x_{0}x_{2}\in E(G)\) such that \(x_{0}\) and \(x_{1}\) have degree 2 and \(x_{2}\) has degree d, where \(d\geq3\). Now there arise two subcases.

Subcase 2.1: \(x_{1}\) and \(x_{2}\) are nonadjacent vertices in G. Then we construct a new graph H by deleting a vertex \(x_{0}\) and adding an edge between \(x_{1}\) and \(x_{2}\), that is, \(H=G-x_{0}+x_{1}x_{2} \in X(n-1,t)\). Then by inductive assumption we have

$$\begin{aligned} \chi_{\alpha}(G) =&\chi_{\alpha}(H)+(d+2)^{\alpha}+4^{\alpha }-(d+2)^{\alpha} \\ \geq& F(n-1,t)+4^{\alpha}\\ =&F(n,t)+2tn^{\alpha}-2t(n+1)^{\alpha}+(n-2t-2) (n-1)^{\alpha }-(n-2t-1)n^{\alpha}+4^{\alpha}\\ =&F(n,t)+2t\bigl(n^{\alpha}-(n-1)^{\alpha}\bigr)+(n-2t-2) \bigl((n-1)^{\alpha}-n^{\alpha}\bigr)+4^{\alpha}-n^{\alpha} \\ >&F(n,t). \end{aligned}$$

The last inequality holds if \(n\geq7\). If \(n=6\), then \(t\leq2\), and we easily show the inequality.

Subcase 2.2: \(x_{1}\) and \(x_{2}\) are adjacent in G. Then we construct a new graph H by deleting two vertices \(x_{0}\) and \(x_{1}\), that is, \(H=G-x_{0}-x_{1}\in X(n-2,t-1)\). Let the set of neighbors of \(x_{2}\) be \(N_{G}(x_{2})\setminus\{x_{0},x_{1}\}=\{u_{1},u_{2},\dots,u_{d-2}\}\), where all \(u_{1},u_{2},\dots,u_{d-2}\) are nonpendent vertices. By inductive assumption and using Lemmas 2.3 and 2.5, we obtain

$$\begin{aligned} \chi_{\alpha}(G) =&\chi_{\alpha}(H)+4^{\alpha}+2(d+2)^{\alpha}+ \sum_{i=1}^{d-2} \bigl(\bigl(d+d_{G}(u_{i}) \bigr)^{\alpha}-\bigl(d-2+d_{G}(u_{i}) \bigr)^{\alpha}\bigr) \\ \geq&F(n-2,t-1)+4^{\alpha}+2(d+2)^{\alpha}+\sum _{i=1}^{d-2} \bigl(\bigl(d+d_{G}(u_{i}) \bigr)^{\alpha}-\bigl(d-2+d_{G}(u_{i}) \bigr)^{\alpha}\bigr) \\ =&F(n,t)+2(t-1) (n-1)^{\alpha}-2t(n+1)^{\alpha}+(n-2t-1) (n-2)^{\alpha}-(n-2t-1)n^{\alpha}\\ &{}+(t-1)4^{\alpha}-t4^{\alpha}+4^{\alpha}+2(d+2)^{\alpha}+\sum _{i=1}^{d-2} \bigl(\bigl(d+d_{G}(u_{i}) \bigr)^{\alpha}-\bigl(d-2+d_{G}(u_{i}) \bigr)^{\alpha}\bigr) \\ =&F(n,t)+2(t-1) (n-1)^{\alpha}-2t(n+1)^{\alpha}+(n-2t-1) (n-2)^{\alpha}-(n-2t-1)n^{\alpha}\\ &{}+(t-1)4^{\alpha}-t4^{\alpha}+4^{\alpha}+2(d+2)^{\alpha}+(d-2) \bigl((d+2)^{\alpha}-d^{\alpha}\bigr) \\ =&F(n,t)+2(t-1) (n-1)^{\alpha}-2t(n+1)^{\alpha}+(n-2t-1) (n-2)^{\alpha}-(n-2t-1)n^{\alpha}\\ &{}+d(d+2)^{\alpha} -(d-2)d^{\alpha}\\ =&F(n,t)+2(t-1) (n-1)^{\alpha}-2t(n+1)^{\alpha}+(n-2t-1) (n-2)^{\alpha}-(n-2t-1)n^{\alpha}\\ &{}+(n-1) (n+1)^{\alpha}-(n-3) (n-1)^{\alpha}\\ =&F(n,t)+(n-1-2t) \bigl((n+1)^{\alpha}-(n-1)^{\alpha}+(n-2)^{\alpha}-n^{\alpha}\bigr) \\ \geq&F(n,t). \end{aligned}$$

The equality holds if and only if \(d=n-1\), \(2t=n-1\), and \(H\cong X^{0}(n-2,t-1)\). Therefore, we have \(\chi_{\alpha}(G)=F(n,t)\) with equality if and only if \(G\cong X^{0}(n,t)\). □

3 General Randić index

In this section, we find sharp lower bound for the general Randić index of cacti. Let

$$ F_{1}(n,t)=2^{\alpha+1}t(n-1)+t4^{\alpha}+(n-1-2t) (n-1)^{\alpha}, $$

where \(n\geq3\) and \(t\geq0\). First, we give some lemmas that will be helpful in the proof of the main result.

Lemma 3.1

[12]

Let \(n\geq5\) be a positive integer, \(\alpha<0\) be a real number, and \(G\in X(n,0)\). Then

$$ R_{\alpha}(G)\geq(n-1)^{\alpha+1}. $$

Equality holds if and only if \(G\cong X^{0}(n,0)\).

Lemma 3.2

[13]

Let \(n\geq5\) be a positive integer, \(-1\leq\alpha<0\) be a real number, and \(G\in X(n,1)\). Then

$$ R_{\alpha}(G)\geq(n-3) (n-1)^{\alpha}+2^{\alpha}(n-1)^{\alpha}+4^{\alpha}. $$

Equality holds if and only if \(G\cong X^{0}(n,1)\).

Lemma 3.3

Let \(x,d,k\) be positive integers with \(d\geq k\geq1\), \(x\geq2\), and \(\alpha<0\) be a real number. Define

$$ f(x)=x^{\alpha}\bigl(d^{\alpha}-(d-k)^{\alpha}\bigr). $$

Then \(f(x)\) is an increasing function.

Proof

It is easily seen that \(f'(x)=\alpha x^{\alpha-1}(d^{\alpha }-(d-k)^{\alpha})>0\) for \(\alpha<0\). Therefore, \(f(x)\) is an increasing function. □

Lemma 3.4

Let \(x,k\) be positive integers with \(x\geq k\geq1\), and \(-1\leq\alpha <0\) be a real number. Define

$$ f(x)=kx^{\alpha}+2^{\alpha}(x-k)x^{\alpha}-2^{\alpha}(x-k)^{\alpha+1}. $$

Then \(f(x)\) is a decreasing function.

Proof

Let \(g(x)=kx^{\alpha}(1-2^{\alpha})\) for \(x\geq k\geq1\). We get \(g'(x)={\alpha}kx^{\alpha-1}(1-2^{\alpha})<0\). So, \(g(x)\) is a decreasing function.

Let \(h(x)=2^{\alpha}x^{\alpha+1}\) for \(x\geq k\geq1\). Then \(h''(x)=2^{\alpha}(\alpha+1)x^{\alpha-1}<0\) for \(-1\leq\alpha<0\). Hence, \(h(x)-h(x-k)\) is a decreasing function. Note that \(f(x)=g(x)+(h(x)-h(x-k))\). Thus, \(f(x)\) is a decreasing function. □

Lemma 3.5

Let \(x,k\) be positive integers with \(x\geq k\geq1\), and \(-1\leq\alpha <0\) be areal number. Let

$$ f(x)=x^{\alpha+1}-(x-2)^{\alpha+1}. $$

Then \(f(x)\) is a decreasing function.

Proof

Let \(h(x)=x^{\alpha+1}\) for \(x\geq k\geq1\). Then \(h''(x)=\alpha(\alpha +1)x^{\alpha-1}<0\) for \(-1\leq\alpha<0\). But \(f(x)=h(x)-h(x-2)\). Thus \(f(x)\) is a decreasing function. □

Theorem 3.1

Let \(-1\leq\alpha<0\) be a real number, \(n\geq3\) be a positive integer, and \(G\in X(n,t)\). Then

$$ R_{\alpha}(G)\geq F_{1}(n,t). $$
(3.1)

Equality holds if and only if \(G\cong X^{0}(n,t)\).

Proof

We use mathematical induction on n and t. If \(t=0\) or \(t=1\), then inequality (3.1) holds by Lemma 3.1 and Lemma 3.2, respectively. If \(n=5\) and \(t=2\), then there is only one graph \(X^{0}(5,2)\), and the result for \(X^{0}(5,2)\) is trivial. Now if \(n\geq6\) and \(t\geq2\), then we will consider the following two cases.

Case 1: \(G\in X(n,t)\) has at least one pendent vertex.

Let x be a vertex that is adjacent only with vertex y. Then y has degree d with \(2\leq d\leq n-1\). The set of neighbors of y in G is \(N_{G}(y)\setminus\{x\}=\{u_{1},u_{2},u_{3},\dots,u_{d-1}\}\). Without lost of generality, we assume that \(d_{G}(u_{j})\geq2\) for \(k\leq j\leq d-1\), where \(k\geq1\), and the remaining neighboring vertices of y are pendent vertices. If \(k=1\), then G has no pendent vertex. Obviously, we have \(t\leq\lfloor\frac{n-k-1}{2}\rfloor\). If H is a new graph obtained by deleting vertices \(x,u_{1},u_{2},\dots,u_{k-1}\), that is, \(H=G-x-u_{1}-u_{2}-\cdots-u_{k-1}\), then \(H\in X(n-k,t)\). By inductive assumption and using Lemmas 3.3 and 3.4, we obtain

$$\begin{aligned} R_{\alpha}(G) =&R_{\alpha}(H)+kd^{\alpha}+\sum _{j=k}^{d-1} \bigl(\bigl(dd_{G}(u_{j}) \bigr)^{\alpha}-(d-k)^{\alpha}d_{G}(u_{j})^{\alpha} \bigr) \\ \geq&F_{1}(n-k,t)+kd^{\alpha}+\sum _{j=k}^{d-1} \bigl(\bigl(dd_{G}(u_{j}) \bigr)^{\alpha}-(d-k)^{\alpha}d_{G}(u_{j})^{\alpha} \bigr) \\ =&F_{1}(n,t)+2^{\alpha+1}t(n-k+1)^{\alpha}-2^{\alpha+1}t(n-1)^{\alpha }+(n-k-2t-1) (n-k-1)^{\alpha}\\ &{}-(n-2t-1) (n-1)^{\alpha}+kd^{\alpha}+\sum _{j=k}^{d-1} \bigl(\bigl(dd_{G}(u_{j}) \bigr)^{\alpha}-(d-k)^{\alpha}d_{G}(u_{j})^{\alpha} \bigr) \\ \geq& F_{1}(n,t)+2^{\alpha+1}t(n-k+1)^{\alpha}-2^{\alpha+1}t(n-1)^{\alpha }+(n-k-2t-1) (n-k-1)^{\alpha}\\ &{}-(n-2t-1) (n-1)^{\alpha}+kd^{\alpha}+(d-k) \bigl((2d)^{\alpha}-2^{\alpha }(d-k)^{\alpha} \bigr) \\ \geq&F_{1}(n,t)+2^{\alpha+1}t(n-k+1)^{\alpha}-2^{\alpha+1}t(n-1)^{\alpha }+(n-k-2t-1) (n-k-1)^{\alpha}\\ &{}-(n-2t-1) (n-1)^{\alpha}+k(n-1)^{\alpha}+2^{\alpha}(n-1-k) (n-1)^{\alpha }-2^{\alpha}(n+1-k)^{\alpha+1} \\ =&F_{1}(n,t)+(n-1-k-2t) \bigl(1-2^{\alpha}\bigr) \bigl((n-k-1)^{\alpha}-(n-1)^{\alpha } \bigr) \\ \geq&F_{1}(n,t). \end{aligned}$$

The equality holds if and only if \(H\cong X^{0}(n-k,t)\), \(d=n-1\), and \(2t=n-k-1\). Therefore, we have that \(R_{\alpha}(G)=F_{1}(n,t)\) if and only if \(G\cong X^{0}(n,t)\).

Case 2: If G has no pendent vertex, then we consider \(x_{0}x_{1},x_{0}x_{2}\) edges of a cycle such that \(x_{0}\) and \(x_{1}\) are vertices of degree two and \(d_{G}(x_{2})=d\), where \(d\geq3\). Next, we discuss this in the following two subcases.

Subcase 2.1: If \(x_{1}\) and \(x_{2}\) are nonadjacent vertices in G, then H is a new graph obtained by deleting \(x_{0}\) and adding an edge \(x_{1}x_{2}\), that is, \(H=G-x_{0}+x_{1}x_{2} \in X(n-1,t)\). Therefore, we get

$$\begin{aligned} R_{\alpha}(G) =&R_{\alpha}(H)+(2d)^{\alpha}+4^{\alpha}-(2d)^{\alpha} \\ \geq& F_{1}(n-1,t)+4^{\alpha}\\ =&F_{1}(n,t)+2^{\alpha+1}t(n-2)^{\alpha}-2^{\alpha+1}t(n-1)^{\alpha }+(n-2t-2) (n-2)^{\alpha}\\ &{}-(n-2t-1) (n-1)^{\alpha}+4^{\alpha}\\ =&F_{1}(n,t)+2^{\alpha+1}t\bigl((n-2)^{\alpha}-(n-1)^{\alpha}\bigr)+(n-2t-2) \bigl((n-1)^{\alpha}-n^{\alpha}\bigr)\\ &{}+4^{\alpha}-(n-1)^{\alpha} \\ >&F_{1}(n,t). \end{aligned}$$

The last inequality holds if \(n\geq7\). If \(n=6\), then \(t\leq2\), and we easily show the inequality.

Subcase 2.2: If \(x_{1}\) and \(x_{2}\) are connected by an edge, then \(H=G-x_{0}-x_{1}\in X(n-2,t-1)\). Let the set of neighbors of \(x_{2}\) be \(N_{G}(x_{2})\setminus\{x_{0},x_{1}\}=\{u_{1},u_{2},\dots,u_{d-2}\}\), where \(u_{j}\) for \(1\leq j\leq d-2\) are nonpendent vertices. By inductive assumption and using Lemmas 3.3 and 3.5, we obtain

$$\begin{aligned} R_{\alpha}(G) =&R_{\alpha}(H)+4^{\alpha}+2^{\alpha+1}d^{\alpha}+ \sum_{j=1}^{d-2} \bigl(\bigl(dd_{G}(u_{j}) \bigr)^{\alpha}-(d-2)^{\alpha }d_{G}(u_{j})^{\alpha}\bigr) \\ \geq&F_{1}(n-2,t-1)+4^{\alpha}+2^{\alpha+1}d^{\alpha}+ \sum_{j=1}^{d-2} \bigl(\bigl(dd_{G}(u_{j}) \bigr)^{\alpha}-(d-2)^{\alpha}d_{G}(u_{j})^{\alpha}\bigr) \\ =&F_{1}(n,t)+2^{\alpha+1}(t-1) (n-3)^{\alpha}-2^{\alpha+1}t(n-1)^{\alpha}+(n-2t-1) (n-3)^{\alpha}\\ &{}-(n-2t-1) (n-1)^{\alpha}+(t-1)4^{\alpha}-t4^{\alpha}+4^{\alpha}+2^{\alpha+1}d^{\alpha}\\ &{}+ \sum_{j=1}^{d-2} \bigl(\bigl(dd_{G}(u_{j}) \bigr)^{\alpha}-(d-2)^{\alpha}+d_{G}(u_{j})^{\alpha}\bigr) \\ =&F_{1}(n,t)2^{\alpha+1}(t-1) (n-3)^{\alpha}-2^{\alpha+1}t(n-1)^{\alpha}+(n-2t-1) (n-3)^{\alpha}\\ &{}-(n-2t-1) (n-1)^{\alpha}+{}(t-1)4^{\alpha}-t4^{\alpha}+4^{\alpha}+2^{\alpha+1}d^{\alpha}\\ &{}+2^{\alpha}(d-2) \bigl(d^{\alpha}-(d-2)^{\alpha}\bigr) \\ =&F_{1}(n,t)+2^{\alpha+1}(t-1) (n-3)^{\alpha}-2^{\alpha+1}t(n-1)^{\alpha}+(n-2t-1) (n-3)^{\alpha}\\ &{}-(n-2t-1) (n-1)^{\alpha}+2^{\alpha}d^{\alpha+1}-2^{\alpha}(d-2)^{\alpha+1} \\ =&F_{1}(n,t)+2^{\alpha+1}(t-1) (n-3)^{\alpha}-2^{\alpha+1}t(n-1)^{\alpha}+(n-2t-1) (n-3)^{\alpha}\\ &{}-(n-2t-1) (n-1)^{\alpha}+2^{\alpha}(n-1)^{\alpha}-2^{\alpha}(n-3)^{\alpha}\\ =&F_{1}(n,t)+(n-1-2t) \bigl(1-2^{\alpha}\bigr) \bigl((n-3)^{\alpha}-(n-1)^{\alpha}\bigr) \\ \geq&F_{1}(n,t). \end{aligned}$$

The equality holds if and only if \(H\cong X^{0}(n-2,t-1)\), \(d=n-1\), and \(2t=n-1\). Therefore, we have that \(R_{\alpha}(G)=F_{1}(n,t)\) if and only if \(G\cong X^{0}(n,t)\). □

4 Conclusion

In this paper, we determined the sharp lower bounds for the general sum-connectivity index and the general Randić index of cacti with fixed number of cycles for \(-1\leq\alpha<0\). The general sum-connectivity index and general Randić index of cacti for other values of α remains an open problem. Moreover, some topological indices and polynomials are still unknown for cacti with fixed number of cycles and fixed number of pendent vertices.