<s>[INST] Symmetric inequality with three variables including radicals Let $a, b, c$ be positive reals. Show that the following inequality holds:
$$\frac{a}{\sqrt{a^2 + b^2}} + \frac{b}{\sqrt{b^2 + c^2}} + \frac{c}{\sqrt{c^2 + a^2}} \le \frac{3}{\sqrt2}$$
I managed to do the following:
The  ineq. is equivalent to this:
$$\sqrt{\frac{a^2}{a^2 + b^2}} + \sqrt{\frac{b^2}{b^2 + c^2}} + \sqrt{\frac{c^2}{c^2 + a^2}}\le \frac{3}{\sqrt2}$$
Applying Jensen to the concave $f(x) = \sqrt x$ we get:
$$\sum{\sqrt{\frac{a^2}{a^2 + b^2}}} \le 3 \sqrt{\frac{\sum \frac{a^2}{a^2 + b^2}}{3}}$$ with $\sum$ denoting cyclic sums.
It suffices to prove:
$$\sum\frac{a^2}{a^2 + b^2} \le \frac{3}{2}$$
Or:
$$\sum\frac{b^2}{a^2 + b^2} \ge \frac{3}{2}$$
This is where I am stuck. As Michał Miśkiewicz's comment stated, this can't hold, so we must try something else, but I have no ideas. Can someone help?
 [/INST] By C-S
$$\left(\sum_{cyc}\frac{a}{\sqrt{a^2+b^2}}\right)^2\leq\sum_{cyc}\frac{a^2}{(a^2+b^2)(a^2+c^2)}\sum_{cyc}(a^2+c^2)\leq\frac{9}{2},$$
where the last inequality it's
$$9\prod_{cyc}(a^2+b^2)\geq8\sum_{cyc}a^2\sum_{cyc}a^2b^2,$$ which is
$$\sum_{cyc}c^2(a^2-b^2)^2\geq0.$$
Also, we can use Jensen here:
$$\sum_{cyc}\frac{a}{\sqrt{a^2+b^2}}=\sum_{cyc}\left(\frac{a^2+c^2}{2(a^2+b^2+c^2)}\cdot\sqrt{\frac{4a^2(a^2+b^2+c^2)^2}{(a^2+b^2)(a^2+c^2)^2}}\right)\leq$$
$$\leq\sqrt{\sum_{cyc}\left(\frac{a^2+c^2}{2(a^2+b^2+c^2)}\cdot\frac{4a^2(a^2+b^2+c^2)^2}{(a^2+b^2)(a^2+c^2)^2}\right)}=$$
$$=\sqrt{\frac{4(a^2+b^2+c^2)(a^2b^2+a^2c^2+b^2c^2)}{(a^2+b^2)(a^2+c^2)(b^2+c^2)}}\leq\frac{3}{\sqrt2},$$
where the last inequality it's $\sum\limits_{cyc}c^2(a^2-b^2)^2\geq0$ again.
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