Prove that for any positive real numbers $a$ and $b$,$$a^4 + 2a^3b + 2ab^3 + b^4 ≥ 6a^2b^2.$$I tried using Vieta's formula to show the product of the LHS is greater than the RHS, but I don't think I am correct.
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1What does this have to do with Vieta's formula? – Ѕᴀᴀᴅ Jun 26 '20 at 07:48
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1Have you tried AM-GM inequality? If yes, where are you stuck? – Calvin Lin Jun 26 '20 at 07:51
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1I have not tried the AM-GM inequality. I did think of that, but wasn't exactly sure how to start. Thanks for all the answers I see. This helps a lot. – 60minutemen Jun 26 '20 at 07:57
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Direct AM-GM? $ $ – Maximilian Janisch Jun 28 '20 at 15:43
3 Answers
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Rewrite the inequality and employ AM-GM:
$$a^4+a^3b+a^3b+ab^3+ab^3+b^4 \ge 6\sqrt[6]{a^{12}b^{12}} =6a^2b^2$$
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By AM-GM inequality we have $$a^4+b^4\geq 2a^2b^2$$ and $$a^3b+ab^3\geq 2a^2b^2 $$ Adding both the quations after multiplying the second one by $2$ we get the desired inequality.
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You can easily prove it using AM-GM Inequality or Lagrange method. Using AM-GM Inequality,
$$ \frac {a^4 + b^4}{2} \ge a^2.b^2 $$ $$ \frac {a^2 + b^2}{2} \ge a.b$$
$$ \text {Also, you can write }a^3.b+a.b^3 = ab.(a^2+b^2)$$
$$ \text {So, } a^4 + b^4 +2a^3.b+2a.b^3 \ge 6.a^2.b^2$$
Math Lover
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