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While solving one inequality which posted by @arqady-AOPS. (See unsolved inequality.) $$\color{black}{\sqrt{24a^2b+25}+\sqrt{24b^2c+25}+\sqrt{24c^2a+25}\le 21,}$$I arrived at a much simpler, but still nontrivial inequality$$\color{black}{\sqrt{144a^2b+25c^2}+\sqrt{144b^2c+25a^2}+\sqrt{144c^2a+25b^2}\le 39.}$$ where $a,b,c$ are non-negative real numbers and $a+b+c=3.$

It apparently holds, but I haven't found a proof so far.

As @arqady wrote the title of his AOPS post, this kind of cyclic sum of square root inequalities seems nice look but not easy at all.

We can easily check that equality holds at $(a,b,c)=(1,1,1); (2,1,0)$ and its cyclic permutations.

That's why$$\color{black}{\sqrt{144a^2b+25c^2}+\sqrt{144b^2c+25a^2}+\sqrt{144c^2a+25b^2}\le \sqrt{3}\cdot \sqrt{144(a^2b+b^2c+c^2a)+25(a^2+b^2+c^2)}\le 39}$$ doesn't help here.

I hope you can help me prove it. Any ideas and comments are welcome.

Michael Rozenberg
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Anonymous
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1 Answers1

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By C-S $$\sum_{cyc}\sqrt{144a^2b+25c^2}\leq$$ $$\leq\sqrt{\sum_{cyc}\frac{144a^2b+25c^2}{58a^2+28b^2+83c^2+338ab}\sum_{cyc}(58a^2+28b^2+83c^2+338ab)}=$$ $$=39\sqrt{\sum_{cyc}\frac{144a^2b+25c^2}{58a^2+28b^2+83c^2+338ab}}$$ and it's enough to prove that: $$\sum_{cyc}\frac{432a^2b+25c^2(a+b+c)}{58a^2+28b^2+83c^2+338ab}\leq a+b+c,$$ which is obvious by BW: https://www.wolframalpha.com/input?i=x%2By%2Bz-%28432x%5E2y%2B25z%5E2%28x%2By%2Bz%29%29%2F%2858x%5E2%2B28y%5E2%2B83z%5E2%2B338xy%29-%28432y%5E2z%2B25x%5E2%28x%2By%2Bz%29%29%2F%2858y%5E2%2B28z%5E2%2B83x%5E2%2B338yz%29-%28432z%5E2x%2B25y%5E2%28x%2By%2Bz%29%29%2F%2858z%5E2%2B28x%5E2%2B83y%5E2%2B338zx%29%2Cx%3Da%2Cy%3Da%2Bu%2Cz%3Da%2Bv

Michael Rozenberg
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