# Prove $3(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\ge xyz(x+y+z)^3$

if $x,y,z$ are positive real numbers,Prove:$$3(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\ge xyz(x+y+z)^3$$

Additional info:$\sum_{cyc}$ denotes sums over cyclic permutations of the symbols $x,y,z$. I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.

Things I have done: Writing Left hand side in expanded form:$$LHS=3(\sum_{cyc}x^3y^3+\sum_{cyc}xy^4z+3x^2y^2z^2)$$

Right now I don't have good idea were to start(Is expanding and looking for AM-GM is good?).So any hint is appreciated.

• For $x=y=z$ the inequality states $27x^6 \geq 9x^5$ which certainly is not true for all $x>0$. – user133281 Aug 27 '14 at 16:14
• It's likely $(x+y+z)^3$ on LHS. – Quang Hoang Aug 27 '14 at 16:18
• @user133281,you are right.Maybe there is a missing assumption.like $x+y+z=1$? – user2838619 Aug 27 '14 at 16:18
• Or equivalently, what Quang Hoang said. – Bart Michels Aug 27 '14 at 16:19
• I edited it as @QuangHoang suggested it maybe the correct form. – user2838619 Aug 27 '14 at 16:23

The inequality is equivalent to

$$3\sum x^3y^3 +2\sum xy^4z+3x^2y^2z^2\ge 3\sum(xy^3z^2+x^2y^3z) =3\sum xy^3z(x+z)$$

AM-GM gives

$$x^3y^3+xy^4z+x^2y^2z^2\ge 3 x^2y^3z$$ $$y^3z^3+xy^4z+x^2y^2z^2\ge 3xy^3z^2$$ Summing up we have $$2\sum x^3y^3+2\sum xy^4z+6x^2y^2z^2\ge 3\sum (x^2y^3z+xy^3z^2),$$ or $$\sum x^3y^3+\sum xy^4z+3x^2y^2z^2\ge \frac32\sum(x^2y^3z+xy^3z^2).\tag{1}$$ Now, note that $$x^3+y^3\ge xy^2+yx^2.\tag{2}$$ Multiplying (2) with $xyz$ gives $$\sum xy^4z\ge \frac12\sum(x^2y^3z+x^3y^2z)\tag{3}.$$ Multiplying (2) with $z^3$ gives $$\sum (x^3z^3+y^3z^3)\ge \sum(xy^2z^3+x^2yz^3).\tag{4}$$ The inequality follows from (1), (3), and (4).

Note: It may not be a good book to work with after all. And I'm sure there are other non brute-force proofs.

• thanks for posting your solution.and about the book I agree with you.but I should solve it questions anyway.could you recommend any good books about inequalities?I have Cauchy-Schwartz master class,Algebraic inequalities and Inequalities Theorems,Techniques and Selected Problems – user2838619 Aug 27 '14 at 17:29
• I recommend Pham Kim Hung's Secret's in inequalities, volumes 1-2. (Haven't found the time so far to read them completely.) It explains common techniques with a lot of examples and challenging problems. – Bart Michels Aug 27 '14 at 17:31

In case you want to expand your list of basic inequalities, by Holder's inequality $$LHS=(1+1+1)(x^2y+y^2z+z^2x)(zx^2+xy^2+yz^2) \ge \left(\sqrt{x^4yz}+\sqrt{xy^4z}+\sqrt{xyz^4}\right)^3=RHS$$

Since $\let\geq\geqslant\let\leq\leqslant$the inequality is homogenous, we may assume $x+y+z=1$. The inequality can be rewritten as $$3\cdot\sum_{cyc}\frac xy\sum_{cyc}\frac yx\geq\frac1{xyz}.$$ By weighted AM-GM, $$\sum_{cyc}\frac xy\geq(x+y+z)\sqrt[x+y+z]{\frac1{y^xx^zz^y}}.$$ Combining a similar inequality for the other factor yields $$LHS\geq3\frac{x^xy^yz^z}{xyz},$$ so it suffices to prove $3x^xy^yz^z\geq1$, which is true by Jensen's inequality in the form $$x\log x+y\log y+z\log z\geq3\frac{x+y+z}3\log\frac{x+y+z}3=-\log 3$$ because $x\log x$ is convex on the interval $]0,1]$.

• Surely weighted AM-GM plus Jensen is an overkill for this, but it's only meant to illustrate a beautiful alternative. – Bart Michels Aug 27 '14 at 18:14