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When I entered a test at my school, I stuck this problem (it is also posted here)

Let $x,y,z$ be positive real numbers such that $x+y+z=3$, prove that $$\frac{1}{x+yz}+\frac{1}{y+zx}+\frac{1}{z+xy} \le \frac{9}{2(xy+yz+zx)}.$$

I have tried AM GM $$\frac{1}{x+yz}\le \frac{1}{2\sqrt{xyz}}.$$ But it leads to a wrong inequality $$(xy+yz+zx)^2 \le 9xyz=3xyz(x+y+z).$$ Also, I tried SOS or Vonicu Schur without any success. It seems impossible to prove $$LHS \le \frac{3}{2} \le RHS$$ Could anyone help me with this problem? Thanks alot.

30 Anh Ti 711
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2 Answers2

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Here is an elementary proof that does not require the $pqr$ method (however using it will tremendously speed up the proof). Notice that: $$\frac{1}{x+yz}=\frac{3}{3x+3yz}=\frac{3}{x^2+xy+xz+3yz}.$$ Using this representation and summing over the cyclic terms, we have the following result after some tedious simplification: $$\sum_{cyc}\frac{1}{x+yz}=\sum_{cyc}\frac{3}{x^2+xy+xz+3yz}=\frac{12(x+y+z)^2(xy+yz+zx)}{\prod_{cyc}(x^2+xy+xz+3yz)}=\frac{4(xy+yz+yx)}{\prod_{cyc}(x+yz)}.$$ Therefore, the original inequality simplifies to: $$ 9(x+yz)(y+zx)(z+xy) - 8(xy+yz+xz)^2\geq 0.$$ By expanding the polynomial, one can show that: $$9(x+yz)(y+zx)(z+xy) - 8(xy+yz+xz)^2 - \frac{1}{2}\sum_{cyc}(x-y)^2z^2=9xyz(x^2+y^2+z^2+xyz-4),$$ and it thus suffices to show that $x^2+y^2+z^2+xyz-4\geq 0$. This can be proved by a standard Lagrange Multiplier. The first order condition is: $$ 2x+yz-\lambda = 0\\ 2y+xz-\lambda = 0\\ 2z+xy-\lambda = 0\\ x+y+z=3 $$ And the solutions are $(x,y,z,\lambda)=(1,1,1,3)$ and $(-1,2,2,2)$ (up to a permutation). Clearly, there is only one solution that satisfies the requirement that $x,y,z>0$. By checking the Hessian matrix, we can confirm that $(x^*,y^*,z^*)=(1,1,1)$ is the global minimum under the constraints on $x,y,z$. Therefore: $x^2+y^2+z^2+xyz-4\geq 1^2+1^2+1^2+1-4=0$ as desired. This completes the proof.

Fred Li
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Not a very elegant solution, but works (Schurhead + Triangle notation). Homogenize before expansion: $$0\leq\frac{9}{2 (x y+x z+y z)}-\frac{1}{\frac{1}{3} x (x+y+z)+y z}-\frac{1}{\frac{1}{3} y (x+y+z)+x z}-\frac{1}{\frac{1}{3} z (x+y+z)+x y}$$ Then we're left to prove: $$x^4 y^2+14 x^4 y z+x^4 z^2+2 x^3 y^3-10 x^3 y^2 z-10 x^3 y z^2+2 x^3 z^3+x^2 y^4-$$$$-10 x^2 y^3 z+6 x^2 y^2 z^2-10 x^2 y z^3+x^2 z^4+14 x y^4 z-10 x y^3 z^2-10 x y^2 z^3+$$$$+14 x y z^4+y^4 z^2+2 y^3 z^3+y^2 z^4\geq0$$

Use the triangle notation: enter image description here

Then we notice the Schur pattern around the central 6, subtract $$2xyz(x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y))$$ Then we're left with: enter image description here

Which can be Cleared by Muirhead since it's $ S[4, 2, 0] + S[3, 3, 0] + 6 S[4, 1, 1] \geq 8 S[3, 2, 1]$

asomog
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