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Let $a,b \in \mathbb{R}.$ Show that $a^4+b^4+8\ge 8ab.$

The question is from the inequalities section of An Excursion in mathematics by Bhaskaraycharya Pratisthanan. My heuristics include using the AM-GM inequality. I am unable to design the problem to proceed further.
Please provide me some hints.

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You're on right path just a small trick using AM-GM Write $$a^4+b^4+8=a^4+b^4+4+4$$ And then apply AM-GM $$\frac{a^4+b^4+4+4}{4}\ge \sqrt[4]{16a^4b^4}$$

$$\frac{a^4+b^4+4+4}{4}\ge 2ab$$

$${a^4+b^4+8}\ge 8ab$$

Or You can just SOS it $$(a^2-b^2)^2\ge0\implies a^4+b^4-2a^2b^2\ge0$$

$$2(ab-2)^2\ge0\implies2a^2b^2+8-8ab\ge0$$ Add above two inequalities to get $$ a^4+b^4+8\ge8ab$$

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  • $\begingroup$ You may also add to your answer how the inequality can be obtained from the Rearrangement Inequality. I think it will help the OP more. $\endgroup$
    – user170039
    Nov 8 '14 at 16:41
  • $\begingroup$ @user170039 how can we obtain it from rearrangement inequality ? $\endgroup$
    – r9m
    Nov 9 '14 at 21:43
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    $\begingroup$ @r9m: The AM-GM Inequality can be proved from Rearrangement Inequality. $\endgroup$
    – user170039
    Nov 10 '14 at 3:05
  • $\begingroup$ @user170039 thanks for clearing it ! :) $\endgroup$
    – r9m
    Nov 10 '14 at 4:24
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We have $$\frac{a^4+b^4}{2}\geq a^2b^2$$ by using AM-GM inequality, therefore we will obtain $$a^4+b^4+8-8ab\geq 2a^2b^2+8-8ab=2(ab-2)^2\geq 0$$

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    $\begingroup$ Do you ever look at your answers and think "that could look a lot better if I had bothered to make it pretty". $\endgroup$
    – Sarah
    Nov 17 '14 at 21:34
  • $\begingroup$ i try to find the answer as fast as i can Sarah $\endgroup$ Nov 17 '14 at 21:35
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    $\begingroup$ I think you would be doing this community a great favour by later revising your answer and making it high quality. $\endgroup$
    – Sarah
    Nov 17 '14 at 21:36
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    $\begingroup$ Proper grammar. Good spacing between formulas. Look at other answers as an example. $\endgroup$
    – Sarah
    Nov 17 '14 at 21:39
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    $\begingroup$ I do not know if you are aware @Dr.SonnhardGraubner but several users have been complaining about the quality of your posts. Posting answers is fine but the ultimate point of SE is to provide high quality questions and answers. Your contribution is welcome as long as you are making an effort to participate. You will find that you will not have as much negative feedback if your answers are well written and formatted as Sarah suggested. $\endgroup$ Nov 17 '14 at 21:43
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Another way to look at it is to consider the following. $$(a^2-b^2)^2\geq0$$ Therefore

$$ a^4+b^4\geq2a^2b^2$$

Then from there you proceed the same way as Dr.Sonnhard did..

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