Prove that $x^2+xy^2+xyz^2 \ge 4xyz-4$ for postive real $x,y,z$.

I tried AM-GM but failed. I also can't apply Cauchy-Schwarz. I think we have to change this inequality to apply any well-known inequality. Please help me.



Now using AM-GM inequality, $$\dfrac{x^2+2\cdot\dfrac{xy^2}2+4\cdot\dfrac{xyz^2}4+4}{1+2+4+1}\ge\sqrt[1+2+4+1]{x^2\cdot\left(\cdot\dfrac{xy^2}2\right)^2\cdot\left(\dfrac{xyz^2}4\right)^4\cdot4}$$

Here is how I've identified the coefficients:

let $$x^2+xy^2+xyz^2+4=a\cdot\dfrac{x^2}a+b\cdot\dfrac{xy^2}b+c\cdot\dfrac{xyz^2}c+d\cdot\dfrac4d$$

Now by AM-GM, $$\dfrac{a\cdot\dfrac{x^2}a+b\cdot\dfrac{xy^2}b+c\cdot\dfrac{xyz^2}c+d\cdot\dfrac4d}{a+b+c+d}\ge?$$

Compare the exponents of $x,y,z$ to find


Choose $a=1$


$$x^2+xy^2+xyz^2-4xyz+4 = x^2+xy^2-4xy+4+xy(z-2)^2 = $$ $$x^2+x(y-2)^2+xy(z-2)^2-4x+4 =(x-2)^2+x(y-2)^2+xy(z-2)^2\geq 0.$$

  • 1
    $\begingroup$ Shouldn't we have $x^2 + x(y - 2)^2 + \color{red}{xy}(z - 2)^2 - 4x + 4 = (x - 2)^2 + x(y - 2)^2 + xy(z - 2)^2$? $\endgroup$ – N. F. Taussig Sep 11 '18 at 8:22

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