Inequality Proof: x/sqrt(2y^2+5) + y/sqrt(2x^2+5) <= 2/sqrt(7)

[anemone]

Here is this week's POTW:

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Let $0\le x \le 1$ and $0\le y \le 1$ . Prove the inequality

\(\displaystyle \frac{x}{\sqrt{2y^2+5}}+\frac{y}{\sqrt{2x^2+5}}\le \frac{2}{\sqrt{7}}.\)

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[anemone]

Congratulations to kaliprasad for his correct solution::)

You can find the proposed solution as follows:

Spoiler

Since $x^2\le 1$, we have $2y^2+5\ge 2y^2+2x^2+3.$

Similarly, we have $2x^2+5\ge 2x^2+2y^2+3$ and it follows that

\(\displaystyle \frac{x}{\sqrt{2y^2+5}}+\frac{y}{\sqrt{2x^2+5}}\le \frac{x+y}{\sqrt{2y^2+2x^2+3}}\)

Therefore, it suffices to show that $\sqrt{7}(x+y)\le 2\sqrt{2y^2+2x^2+3}$. Squaring and rearranging the terms, this is equivalent to $12xy\le (x-y)^2+12$, which is certainly true since $12xy\le 12$ and $(x-y)^2+12\ge 12$.

The result follows. Equality holds if and only if $x=y=1.$