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][1]), 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][2] 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 Gamma$(\lambda/2,2)$ and 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][3], 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. 

One may also note that the lower bound proposed by ofer zeitouni is smaller than the one proposed by axk, at least when $d=1$ and $\lambda_1=1$. 


  [1]: http://ejp.ejpecp.org/article/view/371
  [2]: http://arxiv.org/abs/1501.06599
  [3]: http://www.emis.de/journals/JIPAM/images/157_05_JIPAM/157_05.pdf