For $x+y+z=3,$ prove $\frac{1}{x^2+4}+\frac{1}{y^2+4}+\frac{1}{z^2+4}\le \frac{3}{5}.$

Question

Let $x,y,z\ge 0: x+y+z=3.$ Prove that$$\frac{1}{x^2+4}+\frac{1}{y^2+4}+\frac{1}{z^2+4}\le \frac{3}{5}.$$ Here is just my thought progress. I set $0\le xy+yz+zx=q\le 3; 0\le r=xyz\le 1.$ After full expanding, I get $$7q^2-16q+3r^2-42r+24\ge 0.$$ From now, I tried to use $pqr$ method but failed.

I hope you give me a advice to continue my approach. By the way, if you found others idea, please post it.

Thank you very much.

Answer 1

Let $f(x)=\frac{1}{x^2+4}.$

Thus, $$f''(x)=\frac{6x^2-8}{(x^2+4)^3}\leq0,$$ for $0\leq x\leq\frac{x+y+z}{3}=1.$

Thus, by the Vasc's LCF Theorem it's enough to prove our inequality for equality case of two variables.

Let $y=x$ and $z=3-2x$, where $0\leq0\leq\frac{3}{2}.$

Thus, it's enough to prove that: $$\frac{2}{x^2+4}+\frac{1}{(3-2x)^2+5}\leq\frac{3}{5}$$ or $$(x-1)^2(2x^2-2x+1)\geq0$$ and we are done.

About LCF see here: https://cut-the-knot.org/arithmetic/algebra/VasileCirtoaje.pdf p.143

Answer 2

Proof.

I continued the original proposer's approach.

Indeed, we will prove a quadratic polynomial of $r$ $$f(r)=3r^2-42r+7q^2-16q+24\ge 0.$$ Notice that $f'(r)=6r-42<0,$ hence $f(r)$ is decreasing. We can use $$r \le \dfrac{q^2(p^2-q)}{2p(2p^2-3q)}=\dfrac{q^2(9-q)}{18(6-q)}.\tag {*}$$ Id est, it's enough to prove $$7q^2-16q+3\left(\dfrac{q^2(9-q)}{18(6-q)}\right)^2-42\dfrac{q^2(9-q)}{18(6-q)}+24\ge 0,$$which is equivalent to $$\frac{(q - 3) (q^5 - 15 q^4 + 540 q^3 - 5400 q^2 + 20736 q - 31104)}{108 (q - 6)^2}\ge 0.$$ The last inequality is true for all $0\le q\le 3.$

We end proof here. Equality holds at $x=y=z=1.$

Remark. The inequality $(*)$ can be solved by AM-GM based on the following indentity$$(a-b)^2(b-c)^2(c-a)^2=p^2q^2-4q^3+2p(9q-2p^2)r-27r^2.$$

Answer 3

Another way.

Let $z=\max\{x,y,z\}$ and $f(x,y,z)=\frac{3}{5}-\sum\limits_{cyc}\frac{1}{x^2+4}$.

Thus, $z\geq1$, $x+y\leq2$ and by AM-GM $$f(x,y,z)-f\left(\frac{x+y}{2},\frac{x+y}{2},z\right)=\frac{8}{(x+y)^2+14}-\frac{1}{x^2+4}-\frac{1}{y^2+4}=$$ $$=\frac{(x-y)^2(8-x^2-4xy-y^2)}{((x+y)^2+16)(x^2+4)(y^2+4)}=\frac{(x-y)^2(8-(x+y)^2-2xy)}{((x+y)^2+16)(x^2+4)(y^2+4)}\geq$$ $$\geq\frac{(x-y)^2\left(8-(x+y)^2-\frac{(x+y)^2}{2}\right)}{((x+y)^2+16)(x^2+4)(y^2+4)}\geq\frac{2(x-y)^2}{((x+y)^2+16)(x^2+4)(y^2+4)}\geq0.$$ Thus, it's enough to prove our inequality for equality case of two variables, which for $y=x$ gives $$(x-1)^2(2x^2-2x+1)\geq0.$$

Answer 4

This is a case of total symmetry around the line $L\to p = \lambda(1,1,1)'$. In cases like that we can choose a plane $P$ which contains $L$ and study to solve the proposed problem. Taking $P\to p\cdot \vec n=0$ suffices to choose $\vec n = (1,-1,0)'$ or $x-y=0$. so we follow with: prove

$$ \cases{ \frac{2}{x^2+4}+\frac{1}{z^2+4}\le \frac 35\\ 2x+z = 3 } $$

which is trivially solved as $\max_{x,z}\frac{2}{x^2+4}+\frac{1}{z^2+4}$ s. t $\cases{x\ge 0\\ z\ge 0\\ 2x+z = 3}$ .