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$x,y,z$ are positive real numbers such that $$x^2+y^2+z^2=1$$ Prove that $\dfrac{x^2}{1+2yz}+\dfrac{y^2}{1+2xz}+\dfrac{z^2}{1+2xy} \geqslant \dfrac{3}{5}$.Again, I try with Engel form of Cauchy inequality...

chaos
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4 Answers4

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$$Z=\dfrac{x^2}{1+2yz}+\dfrac{y^2}{1+2xz}+\dfrac{z^2}{1+2xy} $$

By Cauchy-Schwarz: $$\left(\dfrac{x^2}{1+2yz}+\dfrac{y^2}{1+2xz}+\dfrac{z^2}{1+2xy} \right)(x^2(1+2yz)+y^2(1+2xz)+z^2(1+2xy))\geq(x^2+y^2+z^2)^2=1$$ So: $$Z\geq\frac1{x^2+y^2+z^2+2xyz(x+y+z)}=\frac1{1+2xyz(x+y+z)}$$ Now $x+y+z\geq3(xyz)^{1/3}$, but also $1=x^2+y^2+z^2\geq3(xyz)^{2/3}\implies (xyz)^{4/3}\leq1/{9}$

So: $$Z\geq\frac1{1+2xyz(x+y+z)}\geq\frac1{1+6(xyz)^{4/3}}\geq\frac1{1+6(1/9)}=\frac35$$

Daniel Fischer
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RE60K
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Hint: By AM-GM: $$\frac{x^2}{1+2yz} \ge \frac{x^2}{1+y^2+z^2} = \frac{x^2}{2-x^2}$$

Now apply Jensen to the convex function $t \mapsto \dfrac{t^2}{2-t^2}$ to conclude...

Macavity
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Approaching the problem through Engel's form of Cauchy Schwarz-

$$\dfrac{x^4}{x^2+2x^2yz}+\dfrac{y^4}{y^2+2xy^2z}+\dfrac{z^4}{z^2+2xyz^2} \geqslant \dfrac{(x^2+y^2+z^2)^2}{x^2+y^2+z^2+2xyz(x+y+z)}$$

Now, given condition reduces the inequality to - $$\dfrac{x^4}{x^2+2x^2yz}+\dfrac{y^4}{y^2+2xy^2z}+\dfrac{z^4}{z^2+2xyz^2} \geqslant \dfrac{1}{1+2xyz(x+y+z)}$$

If we can prove $\dfrac{1}{1+2xyz(x+y+z)} \geq \frac35$ we are done. It can be rewritten as $\space 5\geq 3 + 6xyz(x+y+z) \\ \iff 1\geq 3xyz(x+y+z)\\ \iff (x^2+y^2+z^2)^2\geq 3xyz(x+y+z) \\ \iff x^4+y^4+z^4 +2x^2y^2+2y^2z^2 +2z^2x^2 \geq 3xyz(x+y+z) $

The last inequality is true as from AM-GM

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

Soham
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Last time I posted similar comments and got deleted so I'm not sure I want to do it again. If someone is going to delete this post please tell me which rule I'm violating.

These types of symmetric inequalities with homogenous condition can almost always be solved using "brute force" by making it equivalents to homogeneous polynomial inequality. It's not pretty but practical. It's like using analytical geometry to approach plane geometry problems.

In this question all variables are positive without specified.

Back to your question, let's define $\sum_{sym}$ to be symmetric sum, for example, $$\sum_{sym}xy=xy+yz+zx$$ $$\sum_{sym}x^2y=x^2y+xy^2+y^2z+yz^2+z^2x+zx^2$$ Now a trivial lemma
$x^ny+y^nx<=x^{n+1}+y^{n+1}$, this is true because you can factorized it into $(x^n-y^n)(x-y)\geq 0$ and it's either ++ or --.
Use this and AM-GM inequality, try to prove the following yourself $$2\sum_{sym}x^6\geq \sum_{sym}x^5y\geq\sum_{sym}x^4y^2\geq 2\sum_{sym}x^3y^3\geq xyz\sum_{sym}x^2y\geq6x^2y^2z^2$$ and $$xyz\sum_{sym}x^3\geq 3x^2y^2z^2$$

The original inequality, replacing 1 by $x^2+y^2+z^2$, by making left hand side common denominator, becomes $$\frac{\sum_{sym}x^6+2\sum_{sym}x^5y+3\sum_{sym}x^4y^2+4\sum_{sym}x^3y^3+4xyz\sum_{sym}x^3+2xyz\sum_{sym}x^2y+6x^2y^2z^2}{\sum_{sym}x^6+2\sum_{sym}x^5y+3\sum_{sym}x^4y^2+4\sum_{sym}x^3y^3+6xyz\sum_{sym}x^3+8xyz\sum_{sym}x^2y+14x^2y^2z^2}\geq \frac{3}{5}$$

Which equivalents to polynomial inequality $$2\sum_{sym}x^6+4\sum_{sym}x^5y+6\sum_{sym}x^4y^2+8\sum_{sym}x^3y^3+2xyz\sum_{sym}x^3-14xyz\sum_{sym}x^2y-12x^2y^2z^2\geq 0$$

The last inequality is trivial from the chain of inequalities stated above.