Have $x=2$ and $b,c\to \infty$. Then we need have $\frac {3} {2a^2} \geq \frac {2} {1+a^2}$, equivalent to $3\geq a^2$, and why should this hold? There is something missing in this "strange inequality".

The problem link : http://www.imocompendium.com/othercomp/Med/MedMO99.pdf @mavropvena I doubt this problem like you, but unsure my sense.

So who is at fault you forgot the square root sign? It is \[\frac{3}{2}\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\geq\frac{x}{1+a^2}+\frac{y}{1+b^2}+\frac{z}{1+c^2}.\] By Lagrange Multipliers method, it is enough to check it for $(x,y,z) = (1,1,1)$, $(x,y,z) = (3/2,3/2,0)$, or $(x,y,z) = (3,0,0)$.

Sorry Now I have corrected the problem

Actually, it's very easy and we don't need Lagrange's multipliers. Inequality is equivalent to: ${\frac{1}{2}\sqrt{\sum \frac{1}{a^2} }} \cdot \sum x \ge \sum \frac{x}{1+a^2}$ ${\Leftrightarrow \sum ( \frac{1}{2}\sqrt{\sum \frac{1}{a^2} }}- \frac{1}{1+a^2} ) x \ge 0$ But this obviously holds because: ${ \frac{1}{2}\sqrt{\sum \frac{1}{a^2} }} > \frac{1}{2a} \ge \frac{1}{1+a^2}$

Let $a,b,c\not= 0$ and $x,y,z\in\mathbb{R}^+$ such that $x+y+z=3$. Prove that \[\frac{3}{2}\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\geq\frac{x}{1+a^2}+\frac{y}{1+b^2}+\frac{z}{1+c^2}\]

Just restating for the Resources section.

Here is a slightly different way of proving this nice inequality. First, let $m=\frac{1}{a^2}$, and define $n,k$ similarly. Observe that, $\frac{1}{1+a^2}=\frac{m}{m+1}$. Hence, we can rewrite the inequality as, $$ \frac{3}{2}\sqrt{m+n+k}\geq \frac{xm}{m+1}+\frac{yn}{n+1}+\frac{zk}{k+1}. $$Now, observe that, using AM-GM inequality, $m+1\geq 2\sqrt{m}$, hence, the right-hand-side can be upper bounded via, $$ \frac{xm}{m+1}+\frac{yn}{n+1}+\frac{zk}{k+1} \leq \frac{x\sqrt{m}}{2}+\frac{y\sqrt{n}}{2}+\frac{z\sqrt{k}}{2}. $$Hence, it suffices to show that, $$ 3\sqrt{m+n+k}\geq x\sqrt{m}+y\sqrt{n}+z\sqrt{k} \iff \sqrt{m+n+k}\geq \frac{x}{3}\sqrt{m}+\frac{y}{3}\sqrt{n}+\frac{z}{3}\sqrt{k}. $$We will now use Jensen's inequality. Note that, the map $\phi(t):t\mapsto \sqrt{t}$ is concave, and $\lambda_1=x/3,\lambda_2=y/3,\lambda_3=z/3$ are positive weights, that add up to 1; therefore Jensen's inequality establishes, $$ \lambda_1\phi(m)+\lambda_2\phi(n)+\lambda_3\phi(k)\leq \phi(\lambda_1 m + \lambda_2 n + \lambda_3 k)=\sqrt{\frac{xm+yn+zk}{3}}. $$Hence, we are left with proving, $$ 3m+3n+3k \geq xm+yn+zk, $$but this is clear, as $x+y+z=3$.