Variation of Nesbitt Inequality with the geometric mean

Question

The problem is the following:

Prove the inequality: $ \frac{\sqrt{pq}}{p + q + 2r}+\frac{\sqrt{qr}}{q + r+2p}+\frac{\sqrt{pr}}{p + r+2q}≤3/4 $ for $p, q, r>0$ real numbers.

I could prove a weaker inequality using the rearrangement inequality, but i haven't been able to solve this inequality. If you substitute $\sqrt{pq}$ with $\frac{p+q}{2}$ you get the famous Nesbitt inequality.

Any help would be appreciated. Thanks in advance!

Answer 1

By AM-GM and C-S we obtain: $$\sum_{cyc}\frac{\sqrt{pq}}{p+q+2r}\leq\frac{1}{4}\sum_{cyc}\frac{(\sqrt{p}+\sqrt{q})^2}{p+q+2r}\leq\frac{1}{4}\sum_{cyc}\left(\frac{(\sqrt{p})^2}{p+r}+\frac{\sqrt{q})^2}{q+r}\right)=$$ $$=\frac{1}{4}\sum_{cyc}\left(\frac{p}{p+r}+\frac{q}{q+r}\right)=\frac{1}{4}\sum_{cyc}\left(\frac{p}{p+r}+\frac{r}{r+p}\right)=\frac{3}{4}.$$ I used the following C-S: $$\frac{a^2}{b}+\frac{c^2}{d}\geq\frac{(a+c)^2}{b+d},$$ where $b$ and $d$ are positives.

Here $a=\sqrt{p},$ $b=p+r,$ $c=\sqrt{q}$ and $d=q+r.$