# Using arithmetic mean>geometric mean

Prove that if a,b.c are distinct positive integers that $$a^4+b^4+c^4>abc(a+b+c)$$ My attempt:

I used the inequality A.M>G.M to get two inequalities First inequality $$\frac{a^4+b^4+c^4}{3} > \sqrt[3]{a^4b^4c^4}$$ or
$$\frac{a^4+b^4+c^4}{3} > abc \sqrt[3]{abc}$$ or new -- first inequality $$\frac{a^4+b^4+c^4}{3abc} > \sqrt[3]{abc}$$

second inequality: $$\frac{a+b+c}{3} > \sqrt[3]{abc}$$ I am seeing the numbers in the required equation variables here but am not able to manipulate these to get the inequality I want?? Please direct me on which step should I take after this??

• can you use $\LaTeX$ please? – Dr. Sonnhard Graubner Oct 14 '15 at 16:36
• Sorry,Dr sonnhard either it is not available in smartphone or I don't how to use it!! – Freelancer Oct 14 '15 at 16:40
• You can ask the part which is very unclear and shabby so that I will edit the question – Freelancer Oct 14 '15 at 16:41
• @null I think you took ABC up in the final part (after so many or)of my first inequality !! Bring it down!! – Freelancer Oct 14 '15 at 16:49
• @Freelancer, the original was /3×a×b×c which does not suggest that abc is part of the denominator. I moved it down to the denominator as you requested, at least in my suggested edit. – null Oct 14 '15 at 16:54

an other way $$2a^4+b^4+c^4\geq 4\sqrt[4]{a^8b^4c^4}=4a^2bc$$ $$2b^4+a^4+c^4\geq 4\sqrt[4]{b^8a^4c^4}=4b^2ac$$ $$2c^4+a^4+b^4\geq 4\sqrt[4]{c^8a^4b^4}=4c^2ab$$ adding this ineqalitis we get desired one

You have to apply it repeatedly on different levels.

First, $(a^4+b^4)/2\ge a^2b^2$. Also, $(a^4+c^4)/2\ge a^2c^2$. Now apply the AM-GM to these two.

$${(a^4+b^4)/2+(a^4+c^4)/2\over2}\ge\sqrt{a^2b^2\cdot a^2c^2}=a^2bc=abc\cdot a$$ Now write two other similar inequalities with $a,b,c$ in different order, and add them all together.

• Thnx,got it!! But I thought that in my attempt( given in question )I was going in the right direction as I was getting two inequalities which had the terms I required and one side of the inequality was the same in both of these so maybe we can get the solution from the inequalities I wrote can you tell me what to do with them?? – Freelancer Oct 14 '15 at 16:57
• Maybe you could use those, but I don't see how. – Ivan Neretin Oct 14 '15 at 17:06

Here is a slightly more general approach to construct the right AM-GM. Note, for non-negative $x,y$ s.t. $x+y\le 1$, $$x a^4+yb^4+(1-x-y)c^4\ge a^{4x}b^{4y}c^{4-4x-4y}$$ Clearly we are done if we have $4x=2,4y=1\implies 4-4x-4y=1$, so $x=\frac12, y=\frac14$ and what you need is to sum $$\frac12a^4+\frac14b^4+\frac14c^4\ge a^2bc$$

• I couldn't understand how you said x+y<1 ## the only thing I know here is (x+y)/2>√xy – Freelancer Oct 15 '15 at 1:56
• $x, y, 1-X-y$ are weights we wish to determine. So we want them all to be in $(0,1)$. – Macavity Oct 15 '15 at 2:02
• Wht are weights?? I don't understand what you are talking about can you explain this a bit more?? – Freelancer Oct 15 '15 at 2:43
• Check the weighted form of AM-GM inequality at en.m.wikipedia.org/wiki/… – Macavity Oct 15 '15 at 3:53

Late answer since it was tagged as duplicate there and this way of solving is neither here nor there.

$$\begin{eqnarray*} abc(a+b+c) & \stackrel{GM-AM}{\leq} & \left(\frac{a+b+c}{3}\right)^3(a+b+c)\\ & = & 3\left(\frac{a+b+c}{3}\right)^4 \\ & \stackrel{x^4 \; is \; convex}{\leq} & 3\cdot \frac{1}{3}(a^4+b^4+c^4) = a^4+b^4+c^4 \end{eqnarray*}$$

Note that the inequality becomes sharp ("<"), if at least two of the numbers are different.