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Is it possible to use the Karamata inequality to prove the following inequalities to be true ?

$$x^{4}+y^{4}+z^{4}\geq x^{2}yz + xy^{2}z + xyz^{2}$$ $$x^{5}+y^{5}\geq x^{3}y^{2} + x^{2}y^{3}$$ $$x^{4}y + xy^{4}\geq x^{3}y^{2} + x^{2}y^{3}$$

for all $x,y,z \in \mathbb{R}^{+}$

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3 Answers 3

Your third inequality is on page 185 of The Cauchy-Schwarz Master Class by J. Michael Steele. I will paraphrase his argument.

The weighted AM-GM inequality gives $$a^2b^3=(ab^4)^{2/3}(a^4b)^{1/3}\leq {2\over 3}ab^4+{1\over 3}a^4b.$$

If we replace $(a,b)$ in turn by the ordered pairs $(x,y)$ and $(y,x)$, then the sum of the resulting bounds gives us $x^2y^3+y^2x^3\leq xy^4+x^4y$.

I think that your other inequalities can be proven in a similar way. I'm not sure how to use Karamata here.

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As Byron observed all your inequalities follow immediately from AM-GM inequality.

To prove the first one, you can do the following:

$$\frac{x^4+x^4+y^4+z^4}{4} \geq x^2yz \,$$ $$\frac{x^4+y^4+y^4+z^4}{4} \geq xy^2z \,$$ $$\frac{x^4+y^4+z^4+z^4}{4} \geq xyz^2 \,$$

and add them together.

The second one follows from

$$\frac{x^5+x^5+x^5+y^5+y^5}{5} \geq x^3y^2$$ $$\frac{x^5+x^5+y^5+y^5+y^5}{5} \geq x^2y^3$$

while the last one

$$\frac{x^4y+x^4y+xy^4}{3} \geq x^3y^2$$ $$\frac{x^4y+xy^4+xy^4}{3} \geq x^2y^3$$

Actually, all your inequalities are a particular case of the Muirhead's Inequality. Muirhead is similar in idea to Karamata inequality, but I think it is not a consequence of it.... I actually doubt that you can use Karamata inequality here, since you have two sets of variables: $x,y,z$ and the powers.

As for Muirhead, your first inequality is just Muirhead for $[4,0,0] \succeq [2,1,1]$, your second inequality is Muirhead $[5,0] \succeq [3,2]$ and the last is Murihead $[4,1] \succeq [3,2]$.

P.S. Here is a better link for Muirhead Inequality. Muirhead Explained

P.P.S. The solution above is just the standard "Prove this particular case of Muirhead by using AM-GM" approach...

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Thanks for the reply @Byron Schmuland. Could I ask what you mean by a particular case of Muirheads? I know the last two fit with the Proto-Muirhead refered to in the The Cauchy-Schwarz Master Class by J. Michael Steele. –  Roy Mark Robinson Apr 30 '12 at 18:13
    
@RoyMarkRobinson This answer is from N.S., not me. –  Byron Schmuland Apr 30 '12 at 18:17
    
The Murihear inequality says the following: if $[a_1,..,a_n]$ majorizes $[b_1,..,b_n]$ then $\sum_{sym} x_1^{a_1}\cdot...\cdot x_n^{a_n} \geq \sum_{sym} x_1^{b_1}\cdot...\cdot x_n^{b_n}$. –  N. S. Apr 30 '12 at 18:31

Define $f(x)=e^x$ and for an inequality of degree $n$ change variables so that $(a,b,c) = \log (x^n,y^n,z^n)$. Then each assertion is that the left hand side, a sum of values of $f$ at some points, is larger than the sum of $f$ values at another set of points obtained from the first one by a double-stochastic transformation (one whose matrix is non-negative with row and column sums equal to 1). Majorization ordering and double-stochastic ordering are the same so this would be true for any convex function in place of $e^x$. The exponential function case gives the Muirhead inequality of which the three inequalities in the question are instances.

The first inequality is thus $$f(a) + f(b) + f(c) \geq f(a/2 + b/4 + c/4) + f(a/4 + b/2 + c/4) + f(a/4 + b/4 + c/2)$$

and the others can be handled the same way.

Karamata inequality is a synonym for majorization.

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Could it then be said that Muirheads follows directly from Karamata by defining f(x)=e^{x}? –  Roy Mark Robinson Apr 30 '12 at 18:34
    
Yes. (((14char filler material))) –  zyx Apr 30 '12 at 19:02

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