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For positive real numbers $a$, $b$, and $c$ prove that: $$\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}.$$

I let $x=\frac{a}{b}$, $y=\frac{b}{c}$, and $z=\frac{c}{a}$. Then inequality becomes

$$A=\frac{x^2}{1+xz}+\frac{y^2}{1+xy}+\frac{z^2}{1+zy} \ge \frac{3}{2}$$

By Cauchy-Schwarz $$(3+xy+yz+zx)A \ge (x+y+z)^2 \ge 3(xy+yz+zx)$$

What then?

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You can end your way.

Since $xyz=1$, by using your work and AM-GM we obtain: $$2(x+y+z)^2=(x+y+z)^2+(x+y+z)^2\geq9+3(xy+xz+yz),$$ which you want to get.

Also, by C-S $$\sum_{cyc}\frac{a^2}{b^2+bc}=\sum_{cyc}\frac{a^4}{a^2b^2+a^2bc}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b^2+a^2bc)}\geq\frac{3}{2}$$ because the last inequality it's $$\sum_{cyc}(2a^4+a^2b^2-3a^2bc)\geq0,$$ which is true by Muirhead or we can use SOS here: $$\sum_{cyc}(2a^4+a^2b^2-3a^2bc)=\sum_{cyc}(2a^4-2a^2b^2)+\frac{3}{2}\sum_{cyc}(2a^2b^2-2a^2bc)=$$ $$=\sum_{cyc}(a^2-b^2)^2+\frac{3}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$ Done!

There is also the following solution.

By Rearrangement and Nesbitt we obtain: $$\sum_{cyc}\frac{a^2}{b^2+bc}=\sum_{cyc}\left(\frac{a^2}{b+c}\cdot\frac{1}{b}\right)\geq\sum_{cyc}\left(\frac{a^2}{b+c}\cdot\frac{1}{a}\right)=\sum_{cyc}\frac{a}{b+c}\geq\frac{3}{2}$$

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If you let $B=xy+yz+zx$, by AM-GM, $B \ge 3$ and

$$A \ge \frac{3B}{B+3}=\frac{3}{1+3/B} \ge \frac32$$

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