WolfusA wrote: $\definecolor{A}{RGB}{250,120,0}\color{A}\fbox{A3.}$ Assume that $a, b, c$ are positive reals such that $a + b + c = 3$. Prove that $$\definecolor{A}{RGB}{200,0,200}\color{A} \frac{1}{8a^2-18a+11}+\frac{1}{8b^2-18b+11}+\frac{1}{8c^2-18c+11}\le 3.$$ It's true for any reals $a$, $b$ and $c$ such that $a+b+c=3.$

arqady, is your proof by $uvw$?

I would be surprised to see tangent line

Let $f(x)=\frac{1}{8x^2-18x+11}$. If $a,b,c\ge\frac{3}{4}$ then $\sum f(a)\le\sum(2a-1)$. Otherwise $\sum f(a)\le 2f\left(\frac{9}{8}\right)+f\left(\frac{3}{4}\right)\le 3$.

Assume that $a, b, c,d$ are positive reals such that $a + b + c+d= 4$. Prove that $$\frac{1}{7a^2-16a+10}+\frac{1}{7b^2-16b+10}+\frac{1}{7c^2-16c+10}+\frac{1}{7d^2-16d+10}\le 4.$$ WolfusA wrote: $\definecolor{A}{RGB}{250,120,0}\color{A}\fbox{A3.}$ Assume that $a, b, c$ are positive reals such that $a + b + c = 3$. Prove that $$\definecolor{A}{RGB}{200,0,200}\color{A} \frac{1}{8a^2-18a+11}+\frac{1}{8b^2-18b+11}+\frac{1}{8c^2-18c+11}\le 3.$$

Assume that $a, b, c$ are positive reals such that $a + b + c = 3$. Prove that $$\definecolor{A}{RGB}{200,0,200}\color{A} \frac{1}{8a^2-18b+11}+\frac{1}{8b^2-18c+11}+\frac{1}{8c^2-18a+11}\le 3.$$

sqing wrote: Assume that $a, b, c,d$ are positive reals such that $a + b + c+d= 4$. Prove that $$\frac{1}{7a^2-16a+10}+\frac{1}{7b^2-16b+10}+\frac{1}{7c^2-16c+10}+\frac{1}{7d^2-16d+10}\le 4.$$ Let $f(x)=\frac{1}{7x^2-16x+10}$. Also let $a\ge b\ge c\ge d>0$. If $d\ge\frac{11}{14}$ then $\sum f(a)\le\sum (2a-1)=4$. If $\frac{11}{14}>d$ then consider two more cases: - if $1\le c$ and $a\le\frac{9}{7}$ then by Jensen $\sum f(a)\le f(d)+3f\left(\frac{4-d}{3}\right)<4;$ - otherwise $\sum f(a)\le f\left(\frac{11}{14}\right)+f(1)+2f\left(\frac{8}{7}\right)<4.$

Assume that $a, b, c$ are positive reals such that $a + b + c = 3$. Prove that$$\frac{bc}{8a^2-18a+11}+\frac{ca}{8b^2-18b+11}+\frac{ab}{8c^2-18c+11}\le 3 $$$$\frac{c}{8a^2-18a+11}+\frac{a}{8b^2-18b+11}+\frac{b}{8c^2-18c+11}\le 3.$$Assume that $a, b, c,d$ are positive reals such that $a + b + c+d= 4$. Prove that$$\frac{1}{8a^2-18a+11}+\frac{1}{8b^2-18b+11}+\frac{1}{8c^2-18c+11}+\frac{1}{8d^2-18d+11}\le 4 $$https://artofproblemsolving.com/community/c4h1813778p17673562 https://artofproblemsolving.com/community/c6h619669p3699674