Prove that for any real numbers $a,b,c,d\geq \frac{1}{3}$ the following inequality holds: $$\sqrt{\frac{a^6}{b^4+c^3}+\frac{b^6}{c^4+d^3}+\frac{c^6}{d^4+a^3}+\frac{d^6}{a^4+b^3}}\geq \frac{a+b+c+d}{4}$$Equality holds when $a=b=c=d=\frac{1}{3}.$

nAalniaOMliO wrote: Prove that for any real numbers $a,b,c \geq \frac{1}{3}$ the following inequality holds: $$\sqrt{\frac{a^6}{b^4+c^3}+\frac{b^6}{c^4+d^3}+\frac{c^6}{d^4+a^3}+\frac{d^6}{a^4+b^3}}\geq \frac{a+b+c+d}{4}$$D. Zmiaikou Because $$\sqrt{\sum_{cyc}\tfrac{a^6}{b^4+c^3}}\geq\sqrt{\sum_{cyc}\tfrac{a^6}{b^4+3c^4}}\geq\sqrt{\tfrac{\left(\sum\limits_{cyc}a^4\right)^2}{\sum\limits_{cyc}(b^4a^2+3c^4a^2)}}\geq\sqrt{\tfrac{\left(\sum\limits_{cyc}a^4\right)^2}{\tfrac{1}{4}\sum\limits_{cyc}a^2\sum\limits_{cyc}a^4+\tfrac{3}{4}\sum\limits_{cyc}a^2\sum\limits_{cyc}a^4}}=\sqrt{\tfrac{\sum\limits_{cyc}a^4}{\sum\limits_{cyc}a^2}}\geq\sqrt{\tfrac{a^2+b^2+c^2+d^2}{4}}\geq\tfrac{a+b+c+d}{4}.$$

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Asilbek777 wrote:

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"Goals:IMO Perfect score" My congratulations!