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Let $a, b, c$ - real numbers. Prove that $3(a^4+b^4+c^4)+48\ge8(a^2b+b^2c+c^2a)$

user37238
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Golovach Lena
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  • I think is $3(a^4+b^4+c^4)+48\ge 8(a^3b+b^3c+c^3a)$? –  Oct 10 '13 at 10:16
  • I dont thinks so... – Golovach Lena Oct 10 '13 at 10:18
  • this problem from some book? or tell us background? –  Oct 10 '13 at 10:19
  • I kown famous valise(2004) inequality: $$(a^2+b^2+c^2)^2\ge 3(a^3b+b^3c+c^3a)$$ and if and only if $a=b=c$ and $a:b:c=\sin^2{\dfrac{\pi}{7}}:\sin^2{\dfrac{2\pi}{7}}:\sin^2{\dfrac{3\pi}{7}}$ –  Oct 10 '13 at 10:20
  • It is thought to be frm book, but it iss not for sure... – Golovach Lena Oct 10 '13 at 10:20
  • and from Vasile book,have this $3(a^4+b^4+c^4)+4(a^3b+b^3c+c^3a)\ge0$ for all real numbers $a,b,c$.maybe your inequality is wrong. –  Oct 10 '13 at 10:22
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    Hello, welcome to Math.SE. Please read http://meta.math.stackexchange.com/questions/9959/how-to-ask-a-good-question/9960#9960 and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts. – Shobhit Oct 10 '13 at 10:26
  • The conditions of task are right, we need to prove that inequality: 3(a4+b4+c4)+48≥8(a2b+b2c+c2a) – Golovach Lena Oct 10 '13 at 10:30
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    All that is an old-fashioned math. The Mathematica command $$ Minimize[3(a^4 + b^4 + c^4) + 48 - 8(a^2b + ac^2 + b^2*c), {a, b, c}]$$ outputs $${0, {a -> 2, b -> 2, c -> 2}} .$$ – user64494 Oct 10 '13 at 19:42
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    @user64494 Mathematica isn't mathematics! – user1729 Oct 16 '13 at 13:50

1 Answers1

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Consider the following inequality obtained by AM-GM inequality: $$ 2a^4+b^4+16=a^4+a^4+b^4+16\geq 4\sqrt[4]{16a^8b^4}=8ba^2 $$ Writing down similar inequalities for other pairs we get: $$ 2b^4+c^4+16\geq 8cb^2\\ 2c^4+a^4+16\geq 8ac^2 $$ It is enough to sum up all these inequalities and we get the desired inequality.

Arash
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