Proving an inequality using AM-GM

Question

Let $ a,b,c > 0$ such that $ abc = 1$. Prove that $$\frac {ab}{a^2 + b^2 + \sqrt {c}} + \frac {bc}{b^2 + c^2 + \sqrt {a}} + \frac {ca}{c^2 + a^2 + \sqrt {b}}\le 1.$$

$$\frac{ab}{a^2+b^2+\sqrt{c}}=\frac{1}{\frac{a}{b}+\frac{b}{c}+\sqrt{c^3}}$$

(AM-GM): $$\frac{a}{b}+\frac{b}{a}+\sqrt{c^3}\geq 3\sqrt{c} $$

$$\frac{1}{\frac{a}{b}+\frac{b}{a}+\sqrt{c^3}}\leq \frac{1}{3\sqrt{c}}$$

Analogous: $$\frac{1}{\frac{b}{c}+\frac{c}{b}+\sqrt{a^3}} \leq \frac{1}{3\sqrt a}$$

$$\frac{1}{\frac{a}{c}+\frac{c}{a}+\sqrt{b^3}}\leq \frac{1}{3\sqrt b}$$

$$LHS \leq \frac{1}{3}+\frac{1}{3}+\frac{1}{3} =1$$

Am I right?

Answer 1

By AM-GM and C-S we obtain: $$\sum_{cyc}\frac{ab}{a^2+b^2+\sqrt{c}}\leq\sum_{cyc}\frac{ab}{2ab+\sqrt{c^2ab}}=\sum_{cyc}\frac{\sqrt{ab}}{2\sqrt{ab}+c}=$$ $$=\frac{3}{2}-\sum_{cyc}\left(\frac{1}{2}-\frac{\sqrt{ab}}{2\sqrt{ab}+c}\right)=\frac{3}{2}-\sum_{cyc}\frac{c}{2(2\sqrt{ab}+c)}\leq$$ $$\leq\frac{3}{2}-\frac{\left(\sum\limits_{cyc}\sqrt{c}\right)^2}{2\sum\limits_{cyc}(2\sqrt{ab}+c)}=\frac{3}{2}-\frac{1}{2}=1.$$

Answer 2

Here is an easier way to prove the inequality. Just prove:

$$\frac {ab}{a^2 + b^2 + \sqrt {c}} \leq \frac {a+b}{2a + 2b + 2{c}} $$ and you are done! It is equivalent to $$a^2b+ab^2+2\leq a^3+b^3+ a\sqrt{c}+b\sqrt{c}$$ which is easy to see since $2\leq a\sqrt{c}+b\sqrt{c}$ and $ a^2b+ab^2\leq a^3+b^3$ are true by Am-Gm inequality

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Edit to a comment. I tried to find such a number $k$ that a following is true: $$\frac {ab}{a^2 + b^2 + \sqrt {c}} \leq \frac {a^k+b^k}{2a^k + 2b^k + 2c^k} $$ Playing with $a,b,c$ (I put $a=b=x$ and $c=1/x^2$) a find that this is possible only at $k=1$. This is a method I saw when trying to solve: https://artofproblemsolving.com/community/c6h17451p119168