Maths Olympiad Prep

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Algebra Difficulty 6.2 National olympiad Prove it

3. Let positive real numbers a,b,ca, b, c satisfy a+b+c=1a+b+c=1. Prove:
a2+abcab+c+b2+abcbc+a+c2+abcca+b⩽12abc.\frac{\sqrt{a^{2}+a b c}}{a b+c}+\frac{\sqrt{b^{2}+a b c}}{b c+a}+\frac{\sqrt{c^{2}+a b c}}{c a+b} \leqslant \frac{1}{2 \sqrt{a b c}} .

Solution

 3. ∑cyc a2+abcab+c=∑cyca(a+b)(a+c)(a+c)(b+c)=∑cyc aa+ba(a+c)(b+c)2⩽∑cyca+b(a+c)(b+c)2,(a+b)2(a+c)2(b+c)2(a+b+c)⩾4abc∑cyc (a+b)3(a+c).\text { 3. } \begin{array}{l} \sum_{\text {cyc }} \frac{\sqrt{a^{2}+a b c}}{a b+c}=\sum_{\mathrm{cyc}} \frac{\sqrt{a(a+b)(a+c)}}{(a+c)(b+c)}=\sum_{\text {cyc }} a \sqrt{\frac{a+b}{a(a+c)(b+c)^{2}}} \\ \leqslant \sqrt{\sum_{\mathrm{cyc}} \frac{a+b}{(a+c)(b+c)^{2}}}, \\ (a+b)^{2}(a+c)^{2}(b+c)^{2}(a+b+c) \geqslant 4 a b c \sum_{\text {cyc }}(a+b)^{3}(a+c) . \end{array}

From the above, we can conclude that the original inequality holds.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.