A greater and more general lower bound holds:
$$(*)\qquad E f\Big(\sum_{i=1}^d \lambda_i g_i^2\Big) \ge E f(X_\lambda),$$
where $\lambda:=\lambda_1+\dots+\lambda_d$, $X_\lambda$ has the $\chi^2$ distribution with $\lambda$ degrees of freedom (that is, the $\text{Gamma}(\lambda/2,2)$ distribution), and $f$ is any concave function from $[0,\infty)$ to ${\mathbb{R}}$.
Indeed, by a standard induction argument (cf. e.g. the proof of Lemma 5.1.1 in normal domination), conditioning on all the $g_i$'s except one of them and using the convolution property $\text{Gamma}(\alpha_1,\beta)*\text{Gamma}(\alpha_2,\beta)=\text{Gamma}(\alpha_1+\alpha_2,\beta)$ for all positive real $\alpha_1,\alpha_2,\beta$, one sees that without loss of generality $d=1$.
By Corollary 5.7 (with $n=1$, $k=2$, and $I=[0,\infty)$) in dual cones or Corollary 5.8 therein (with $n=1$, $k=2$, and $I={\mathbb{R}}$), it suffices to prove $(*)$ for ($d=1$ and) $f=f_t$, where $t\in[0,\infty)$ and $f_t(x):=-(x-t)_+$ for all real $x$. That is, it suffices to prove that
$$(**)\qquad G_1(t)\le F_1(t)$$
for all real $t\ge0$ and $\lambda\in(0,1)$, where $F_1(t):=F_{1,\lambda}(t):=\int_t^\infty(x-t)_+\,p(x)\,dx$, $G_1(t):=G_{1,\lambda}(t):=\int_t^\infty(x-t)_+\,q(x)\,dx$, and $p$ and $q$ are the $\text{Gamma}(\lambda/2,2)$ and $\text{Gamma}(1/2,2\lambda)$ density functions, respectively.
Introduce also $F(t):=-F_1'(t)=\int_t^\infty p(x)\,dx$ and $G(t):=-G_1'(t)=\int_t^\infty q(x)\,dx$.
Clearly, $F(\infty-)=G(\infty-)=0$. Also, the ratio $q/p$ is incr.-decr. (on $I=[0,\infty)$) -- that is, for some $c\in I$ the ratio $q/p$ is increasing on $[0,c)$ and decreasing on $(c,\infty)$. So, by the l'Hospital-type rule for monotonicity given in Proposition 4.3 in l'Hospital-mono, the ratio $G/F$ is incr.-decr., so that for some $C\in I$ the ratio $G/F$ is increasing on $[0,C)$ and decreasing on $(C,\infty)$. In fact, $C>0$, because in a right neighborhood of $0$ one has $q<p$ and hence $G>F$, whereas $G(0)=F(0)=1$.
By another application of the same l'Hospital-type rule for monotonicity, for some $C_1\in I$ the ratio $G_1/F_1$ is increasing on $[0,C_1)$ and decreasing on $(C_1,\infty)$. In fact, $C_1=0$, because $G_1(0)=F_1(0)=\lambda$ and in a right neighborhood of $0$ one has $G>F$ and hence $G_1<F_1$. That is, $G_1/F_1$ is decreasing on $[0,\infty)$, whence $G_1/F_1\le G_1(0)/F_1(0)=1$ on $[0,\infty)$, so that $(**)$ follows.
Addendum: Details on the reduction to the case $d=1$. As mentioned above, this reduction argument is quite similar to the argument in the proof of Lemma 5.1.1 in normal domination.
Suppose that the inequality in $(*)$ holds for $d=1$. For $i=0,\dots,d$, introduce
$$R_i:=\lambda_1 g_1^2+\dots+\lambda_i g_i^2+X_{\lambda_{i+1}}+\dots+X_{\lambda_d},$$
where the r.v.'s $g_1,\dots,g_d,X_{\lambda_1},\dots,X_{\lambda_d}$ are independent, with $X_{\lambda_i}$ having the $\text{Gamma}(\lambda_i/2,2)$ distribution for each $i=1,\dots,d$.
Let $E_i$ denote the conditional expectation given $g_1,\dots,g_{i-1},
X_{\lambda_{i+1}},\dots,X_{\lambda_d}$. Note that
$R_i-\lambda_i g_i^2=\lambda_1 g_1^2+\dots+\lambda_{i-1} g_{i-1}^2+X_{\lambda_{i+1}}+\dots+X_{\lambda_d}$ is a function of $g_1,\dots,g_{i-1},
X_{\lambda_{i+1}},\dots,X_{\lambda_d}$. Since $f$ is concave, the (random) function $f_i$ given by the formula $f_i(x):=f(R_i-\lambda_i g_i^2+x)$ for all real $x$ is concave, for each $i$. So, by $(*)$ with $d=1$, for all $i=1,\dots,d$,
$$ E_i f(R_i) = E_i f_i(\lambda_i g_i^2) \ge E_i f_i(X_{\lambda_i})
= E_i f(R_{i-1})
$$
and hence $E f(R_i) \ge E f(R_{i-1})$. Thus,
$$E f\Big(\sum_{i=1}^d \lambda_i g_i^2\Big)=E f(R_d) \ge E f(R_0) = E f(X_\lambda),$$
so that $(*)$ follows in general. For the last equality displayed above, one uses the mentioned convolution property of the Gamma family of distributions.
