If $a+b+c=3$ Prove that $a^{2}+b^{2}+c^{2}\geq\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$

Question

Question -

Let $a, b, c$ be positive real numbers such that $a+b+c=3 .$ Prove that $$ a^{2}+b^{2}+c^{2} \geq \frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a} $$

My try -

i tried putting $a+2 = x, b+2=y , c+2=z$

then we get $x+y+z=9$ and after simplification we have to prove that

$3>x/y + y/z + z/x$ which i am not able to prove...

i also tried C-S,Chebyshev,rearrangement..etc but none of them working

any hints ???

thankyou

Answer 1

Here's another way. Notice that $$\frac{2+a}{2+b} = \frac{5a+2b+2c}{2a+5b+2c} = \frac52 -\frac32\cdot\frac{7b+2c}{2a+5b+2c} $$

Also using CS inequality ($\sum $ representing cyclic sums): $$\sum \frac{7b+2c}{2a+5b+2c} \geqslant \frac{\left( \sum (7b+2c)\right)^2}{\sum (7b+2c)(2a+5b+2c)}= \frac{81(a+b+c)^2}{18\sum a^2+21(a+b+c)^2} \\= \frac{729}{18\sum a^2+189}$$

By the above bound, with $x = \sum a^2 \geqslant 3$, it is enough to show $$x \geqslant \frac{15}2- \frac32\cdot\frac{729}{18x + 189} \iff \frac{2(x+6)(x-3)}{2x+21}\geqslant 0$$

Answer 2

Let $c=\min\{a,b,c\}$.

Thus, we need to prove that: $$a^2+b^2+c^2-\frac{(a+b+c)^2}{3}\geq\frac{2+a}{2+b}+\frac{2+b}{2+a}-2+\frac{2+b}{2+c}-\frac{2+b}{2+a}+\frac{2+c}{2+a}-1$$ or $$\frac{2}{3}((a-b)^2+(c-a)(c-b))\geq\frac{(a-b)^2}{(2+a)(2+b)}+\frac{(c-a)(c-b)}{(2+a)(2+c)},$$ which is true because $$\frac{2}{3}>\frac{1}{4}>\frac{1}{(2+a)(2+b)}$$ and $$\frac{2}{3}>\frac{1}{(2+a)(2+c)}.$$