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(to previous question) Given $$x^2+y^2≥2xy$$ Prove that $$a^4+b^4+c^4+d^4≥4abcd.$$ Consider: $$a^4+b^4\geq2\cdot a^2b^2$$ $$c^4+d^4\geq2\cdot c^2d^2$$ So $$a^4+b^4+c^4+d^4 \geq 2(a^2b^2+c^2d^2)$$ \begin{equation*} \geq 2(2abcd) \ \ \ \text{(by part i)} \\ \geq 4abcd \\ \blacksquare \\ \end{equation*}
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Wed, 02 Jun 2021 23:35 GMT