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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?

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  • $\begingroup$ Is any other condition given? $\endgroup$
    – Dr. Sonnhard Graubner
    Apr 8, 2018 at 11:44
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    $\begingroup$ By an odd permutation of variables, $\sum\frac{a^2}{a^2 + b^2} \le \frac{3}{2}$ is equivalent to $\sum\frac{b^2}{a^2 + b^2} \le \frac{3}{2}$, so this inequality cannot hold in general. You need to try some other way. $\endgroup$
    – Michał Miśkiewicz
    Apr 8, 2018 at 11:45
  • $\begingroup$ No there aren't any conditions except the numbers being positive, but we can suppose any condition without loss of generality because the inequality is homogenous $\endgroup$
    – user440781
    Apr 8, 2018 at 11:47
  • $\begingroup$ I edited the question so everybody can see the mistake @MichałMiśkiewicz $\endgroup$
    – user440781
    Apr 8, 2018 at 11:54
  • $\begingroup$ Since the left-hand side is invariant by scaling, you can assume that $(a,b,c)$ lies on the $2$-sphere - that is, $a^2+b^2+c^2=1$. $\endgroup$
    – Arnaud Mortier
    Apr 8, 2018 at 11:59

1 Answer 1

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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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