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let $a,b,c,d$ are positive numbers,and such $$2(a+b+c+d)\ge abcd$$ show that $$a^2+b^2+c^2+d^2\ge abcd$$

My try:if $a,b,c,d\le 16$,then we have $$a^2+b^2+c^2+d^2\ge 4\sqrt{abcd}$$

Andrés E. Caicedo
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  • I like this question. By the way, is this your homework? Where did you get this question? You need to write it, don't you? – mathlove Oct 03 '13 at 12:35

1 Answers1

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You are on the right track. So if, $abcd\le 16,$ then $4\sqrt{abcd}\ge abcd.$ What if $abcd\ge 16?$ Well, then you need another estimate for $a^2+b^2+c^2+d^2.$ The most natural one is $$a^2+b^2+c^2+d^2\ge \frac{(a+b+c+d)^2}{4}\ge \frac{(abcd)^2}{16}\ge abcd. $$

leshik
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