For positive reals $a,b,c$ prove that $$ a^4+b^4+c^4 \ge abc(a+b+c). $$ I tried to pls around trying to reorganize to get AM-GM but i couldn't Thanks for the help in advance.
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1You could take a look at the rearrangement inequality. Alternately, there is probably a Cauchy-Schwartz in there somewhere. – Arthur Sep 11 '19 at 19:24
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How can you say there is cauchy in there... What cues lead you to think that – Tamaghna Chaudhuri Sep 11 '19 at 19:26
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There is always a CS hiding in any inequality. Especially the symmetric ones. That's something I picked up from my IMO days. I never got good at inequalities, but I did pick up that CS was important. – Arthur Sep 11 '19 at 19:28
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Another one: https://math.stackexchange.com/q/1819975/42969. – Martin R Sep 11 '19 at 19:29
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See also https://math.stackexchange.com/questions/1480000/using-arithmetic-meangeometric-mean – Hans Lundmark Sep 11 '19 at 19:42
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I have added an answer based on the convexity of $x^4$ here at the duplicate: https://math.stackexchange.com/questions/1480000/using-arithmetic-meangeometric-mean/3353323#3353323 – trancelocation Sep 11 '19 at 21:22
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Muirhead: $(4,0,0)\gg(2,1,1)$. – Jack D'Aurizio Sep 12 '19 at 14:01
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By AM_GM we get \begin{align*} \frac{2a^4+b^4+c^4}{4} & \geq a^2bc\\ \frac{2b^4+a^4+c^4}{4} & \geq b^2ac\\ \frac{2c^4+b^4+a^4}{4} & \geq c^2ab \end{align*} Now add these to get your inequality.
Anurag A
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Hint
$$\frac{a^4+a^4+b^4+c^4}{4}\geq abca \\ \frac{a^4+b^4+b^4+c^4}{4}\geq abcb \\ \frac{a^4+c^4+b^4+c^4}{4}\geq abcc \\$$
N. S.
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Why should the mean inequalities not work? Geometric mean vs. mean of power 4 $$ abc\le\left(\frac{a^4+b^4+c^4}3\right)^{\frac34} $$and arithmetic mean vs. mean of power 4$$ \frac{a+b+c}3\le\left(\frac{a^4+b^4+c^4}3\right)^{\frac14} $$ implies $$ \frac{abc(a+b+c)}3\le \frac{a^4+b^4+c^4}3. $$
Lutz Lehmann
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@MostafaAyaz : You don't. The mean inequality chain is geometric mean $\le$ arithmetic mean $\le$ quadratic mean $\le$ cubic mean $\le$ quartic mean etc. – Lutz Lehmann Sep 11 '19 at 19:33
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@TamaghnaChaudhuri : This is the mean inequality for the 4th degree power function. In general, for any convex function $f$ you get the Jensen inequality $f(\frac{x_1+...+x_n}n)\le\frac{f(x_1)+...+f(x_n)}n$. Here use $f(x)=x^4$ for the second inequality. Or apply the inequality for arithmetic and quadratic mean twice. – Lutz Lehmann Sep 11 '19 at 19:36
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\begin{eqnarray*} &(x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2 \\&+2(z^2-xy)^2 +2(x^2-yz)^2+2(y^2-zx)^2\geq 0 \end{eqnarray*} now divide by $4$ and rearrange.
Donald Splutterwit
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