Real variables inequality

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High School Math Grades 9-12, Ages 13-18

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#1 h Apr 3, 2025, 12:31 AM

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Let a,b,c\in R then prove that \frac{15}{2}\cdot\frac{a^2+b^2+c^2}{(a+b+c)^2}+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\ge 4

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#2 h Apr 3, 2025, 5:18 AM

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JK1603JK wrote:

Let $a,b,c\in \mathbb{R}$ and $a+b+c \neq 0$ then prove that $\frac{15}{2}\cdot\frac{a^2+b^2+c^2}{(a+b+c)^2}+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\ge 4$

WLOG $a+b+c=5$ . Then the original inequality is equivalent to
$15\sum \frac{a^2}{5^2}+\sum \frac{2ab}{a^2+b^2}=15\sum \frac{a^2}{\left(\sum a\right)^2}+\sum \frac{(a+b)^2}{a^2+b^2}-3\geq 8,$ that is,
$3\sum \frac{a^2}{5}+\sum \frac{(a+b)^2}{a^2+b^2}\geq 11.$
Let
$x=2-a,\quad y=2-b,\quad z=2-c,$ so that
$x+y+z=1.$ Define
$q=xy+yz+zx \quad \text{and} \quad r=xyz.$ Then the inequality is equivalent to proving
$3\sum \frac{(2-x)^2}{5}+\sum \frac{(4-x-y)^2}{(2-x)^2+(2-y)^2}\geq 11.$
According to the Cauchy--Schwarz inequality,
$\sum \frac{(4-x-y)^2}{(2-x)^2+(2-y)^2}\geq \frac{\left(\sum (4-x-y)(6-z)\right)^2}{\sum \left[(2-x)^2+(2-y)^2\right](6-z)^2}.$
Since
$\sum (2-x)^2=9-2q,$ $\sum (4-x-y)(6-z)=56+2q,$ $\sum \left[(2-x)^2+(2-y)^2\right](6-z)^2=560-80q+2q^2+44r,$ it suffices to prove
$\frac{3(9-2q)}{5}+\frac{(56+2q)^2}{560-80q+2q^2+44r}\geq 11.$
Since
$3(x+y+z)xyz\leq (xy+yz+zx)^2,$ we have
$r\leq \frac{q^2}{3},$ and consequently,
$\frac{(56+2q)^2}{560-80q+2q^2+44r}\geq \frac{(56+2q)^2}{560-80q+2q^2+44\cdot \frac{q^2}{3}}.$ Thus, it suffices to prove that
$\frac{3(9-2q)}{5}+\frac{(56+2q)^2}{560-80q+2q^2+44\cdot \frac{q^2}{3}}\geq 11.$
This simplifies to
$(1-3q)q^2\geq 0,$ which is evidently true. Therefore, the original inequality holds.

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