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Question

Cho $a,b,c>0$ . Chứng minh rằng:$ \sqrt{c^2(a^2+b^2)^2+a^2(b^2+c^2)^2+b^2(c^2+a^2)^2}\geq \frac{54(abc)^3}{(a+b+c)^2\sqrt{(ab)^4+(bc)^4+(ca)^4}}$

score 1 · Answer 1

$+\sqrt{\sum a^2(b^2+c^2)^2}\ge\sqrt{\sum 4a^2b^2c^2}=2\sqrt{3}abc$

$+(a+b+c)^2\ge9(\sqrt[3]{abc})^2$

$+\sqrt{(ab)^4+(bc)^4+(ac)^4}\ge\sqrt{3\sqrt[3]{(abc)^8}}=\sqrt{3}\sqrt[6]{(abc)^8}=\sqrt{3}abc.\sqrt[3]{abc}$

$\Rightarrow \sqrt{\sum a^2(b^2+c^2)^2}(a+b+c)^2\sqrt{(ab)^4+(bc)^4+(ac)^4}\ge54(abc)^3$ đpcm

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