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I'm looking for some ideas to solve the following inequality.

Problem. For any non-negative real numbers $a,b,c$ with $a+b+c=ab+bc+ca>0,$ prove that$$\sqrt{\frac{a}{2bc+a+6}}+\sqrt{\frac{b}{2ca+b+6}}+\sqrt{\frac{c}{2ab+c+6}}\ge 1.$$Source: Tran Ngoc Khuong Trang.

Equality holds iff $(a,b,c)=(1,1,1)$ or $(a,b,c)=(2,2,0).$

I've tried to use Holder inequality as $$\left(\sum \sqrt{\frac{a}{2bc+a+6}}\right)^2\cdot \sum a^2(2bc+a+6)(a+k)^3\ge \left[\sum a(a+k)\right]^3$$which saves equality occuring but it leads to wrong inequalities.

Also, I tried to find a human-verified proof without success. It is very interesting to think of a solution using classical inequalities and it is possible to get during a contest time.

Hope you share some ideas to solve this inequality. Thank you for your interest!

Michael Rozenberg
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30 Anh Ti 711
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2 Answers2

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By Holder $$\left(\sum_{cyc}\sqrt{\frac{a}{2bc+a+6}}\right)^2\sum_{cyc}a^2(2bc+a+6)(a+3b+3c)^3\geq\left(\sum_{cyc}(a^2+6ab)\right)^3$$ and it's enough to prove $$\left(\sum_{cyc}(a^2+6ab)\right)^3\geq\sum_{cyc}a^2(2bc+a+6)(a+3b+3c)^3,$$ which after homogenization gives: $$3\sum_{sym}(a^7b^2+13a^6b^3-14a^5b^4)+2abc\sum_{sym}(a^6+25a^5b+62a^4b^2-57a^3b^3)+$$ $$+a^2b^2c^2\sum_{sym}(61a^3-135a^2b+12abc)\geq0,$$ which is true by Schur and Muirhead: $$3\sum_{sym}(a^7b^2+13a^6b^3-14a^5b^4)+2abc\sum_{sym}(a^6+25a^5b+62a^4b^2-57a^3b^3)+$$ $$+a^2b^2c^2\sum_{sym}(61a^3-135a^2b+12abc)\geq$$ $$\geq2abc\sum_{sym}31a^4b^2+a^2b^2c^2\sum_{sym}(61a^3-135a^2b+12abc)\geq$$ $$\geq a^2b^2c^2\sum_{sym}(123a^3-135a^2b+12abc)=$$ $$= a^2b^2c^2\sum_{sym}(12a^3-24a^2b+12abc+111a^3-111a^2b)\geq0.$$

Michael Rozenberg
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Some thoughts.

The so-called isolated fudging works. But it is complicated.

If $abc = 0$, it is easy. In what follows, assume that $a, b, c > 0$.

After homogenization, it suffices to prove that, for all $a, b, c > 0$, $$\sum_{\mathrm{cyc}} \sqrt{\frac{aQ}{2bc+aQ+6Q^2}} \ge 1,$$ where $Q := \frac{ab+bc+ca}{a+b+c}$.

It suffices to prove that, for all $a, b, c > 0$, $$\sqrt{\frac{aQ}{2bc+aQ+6Q^2}} \ge \frac{a^3 + 9a^2(b + c) + 6a(b^2 + c^2) + 23abc}{a^3 + b^3 + c^3 + 15(a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2) + 69abc}. \tag{1}$$ (Note: Taking a cyclic on (1), we get the desired inequality using $$\sum_{\mathrm{cyc}} \frac{a^3 + 9a^2(b + c) + 6a(b^2 + c^2) + 23abc}{a^3 + b^3 + c^3 + 15(a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2) + 69abc} = 1.$$)

WLOG, assume that $a = 1$. We can prove (1) by convexity after substitution $p = b + c, q = bc$.

River Li
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