Cho a, b, c > 0; Chứng minh:
$\frac{a^{4}}{a^{4}+\sqrt[3]{(a^{6}+b^{6})(a^{3}+c^{3})^{2}}}+\frac{b^{4}}{b^{4}+\sqrt[3]{(b^{6}+c^{6})(b^{3}+a^{3})^{2}}}+\frac{c^{4}}{c^{4}+\sqrt[3]{(c^{6}+a^{6})(c^{3}+b^{3})^{2}}}\leq 1$
Cho a, b, c > 0; Chứng minh:
$\frac{a^{4}}{a^{4}+\sqrt[3]{(a^{6}+b^{6})(a^{3}+c^{3})^{2}}}+\frac{b^{4}}{b^{4}+\sqrt[3]{(b^{6}+c^{6})(b^{3}+a^{3})^{2}}}+\frac{c^{4}}{c^{4}+\sqrt[3]{(c^{6}+a^{6})(c^{3}+b^{3})^{2}}}\leq 1$
Cho a, b, c > 0; Chứng minh:
$\frac{a^{4}}{a^{4}+\sqrt[3]{(a^{6}+b^{6})(a^{3}+c^{3})^{2}}}+\frac{b^{4}}{b^{4}+\sqrt[3]{(b^{6}+c^{6})(b^{3}+a^{3})^{2}}}+\frac{c^{4}}{c^{4}+\sqrt[3]{(c^{6}+a^{6})(c^{3}+b^{3})^{2}}}\leq 1$
Áp dụng bđt $Holder$: $(a^6+b^6)(c^3+a^3)(c^3+a^3)\geq (a^2c^2+a^2b^2)^3$
$\Rightarrow\frac{a^4}{a^4+\sqrt[3]{(a^6+b^6)(a^3+c^3)^2}}\leq\frac{a^4}{a^4+a^2c^2+a^2b^2}=\frac{a^2}{\sum a^2}$
$\Rightarrow\sum\frac{a^4}{a^4+\sqrt[3]{(a^6+b^6)(a^3+c^3)^2}}\leq 1$
Dấu "$=$" xảy ra $\Leftrightarrow a=b=c$
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