perhaps, Jensen inequality?

parmenides51 wrote: The sum of non-negative numbers $ x $, $ y $ and $ z $ is $3$. Prove the inequality $$ {1 \over x ^ 2 + y + z} + {1 \over x + y ^ 2 + z} + {1 \over x + y + z ^ 2} \leq 1. $$ Solution. The Cauchy-Schwarz's Inequality gives $(a+b+c)^2\le(a^2+b+c)(1+b+c)$ for any real numbers $a,b,c\ge0$, and thus \begin{align*}&{1 \over x ^ 2 + y + z} + {1 \over x + y ^ 2 + z} + {1 \over x + y + z ^ 2}\\ \leq&\frac{1+y+z}{(x+y+z)^2}+\frac{1+x+z}{(x+y+z)^2}+\frac{1+x+y}{(x+y+z)^2}\\ =&\frac{3+2(x+y+z)}{(x+y+z)^2}=1. \end{align*}As required. $\blacksquare$

parmenides51 wrote: The sum of non-negative numbers $ x $, $ y $ and $ z $ is $3$. Prove the inequality $$ {1 \over x ^ 2 + y + z} + {1 \over x + y ^ 2 + z} + {1 \over x + y + z ^ 2} \leq 1. $$ https://artofproblemsolving.com/community/c6h1775336p11678609 For $a, b, c>0, a+b+c=3$ prove that \[\frac{1}{a^2+b+c+1}+\frac{1}{b^2+c+a+1}+\frac{1}{c^2+a+b+1}\le\frac{3}{4}\]

parmenides51 wrote: The sum of non-negative numbers $ x $, $ y $ and $ z $ is $3$. Prove the inequality $$ {1 \over x ^ 2 + y + z} + {1 \over x + y ^ 2 + z} + {1 \over x + y + z ^ 2} \leq 1. $$ Trivial by $$\forall x\leq 3:\ \frac{1}{x^2-x+3}\leq \frac{-x+4}{9}\iff (x-1)^2 (-x+3)\geq 0$$

sqing wrote: For $a, b, c>0, a+b+c=3$ prove that \[\frac{1}{a^2+b+c+1}+\frac{1}{b^2+c+a+1}+\frac{1}{c^2+a+b+1}\le\frac{3}{4}\] Solution. The Cauchy-Schwarz's Inequality gives $(x+y+z+1)^2\le(x^2+y+z+1)(1+y+z+1)$ for any positive real numbers $x,y,z$, and thus \begin{align*}&\frac{1}{a^2+b+c+1}+\frac{1}{b^2+c+a+1}+\frac{1}{c^2+a+b+1}\\ \leq&\frac{2+b+c}{(a+b+c+1)^2}+\frac{2+c+a}{(a+b+c+1)^2}+\frac{2+a+b}{(a+b+c+1)^2}\\ =&\frac{6+2(a+b+c)}{(a+b+c+1)^2}=\frac{3}{4}. \end{align*}As required. $\blacksquare$

sqing wrote: For $a, b, c>0, a+b+c=3$ prove that \[\frac{1}{a^2+b+c+1}+\frac{1}{b^2+c+a+1}+\frac{1}{c^2+a+b+1}\le\frac{3}{4}\] Also trivial by $$\forall x\leq 4:\ \frac{1}{x^2-x+4}\leq \frac{-x+5}{16}\iff (x-1)^2 (4-x)\geq 0$$

Let $a,b,c$ are positive numbers such that $a+b+c=3.$ Prove that$$\frac{1}{a^2+k(b+c)} +\frac{1}{b^2+k(c+a)}+\frac{1}{c^2+k(a+b)} \le\frac{3}{2k+1}.$$where $1\leq k\leq 2.$

ytChen wrote: parmenides51 wrote: The sum of non-negative numbers $ x $, $ y $ and $ z $ is $3$. Prove the inequality $$ {1 \over x ^ 2 + y + z} + {1 \over x + y ^ 2 + z} + {1 \over x + y + z ^ 2} \leq 1. $$ Solution. The Cauchy-Schwarz's Inequality gives $(a+b+c)^2\le(a^2+b+c)(1+b+c)$ for any real numbers $a,b,c\ge0$, and thus \begin{align*}&{1 \over a ^ 2 + b+ c} + {1 \over a + b^ 2 + c} + {1 \over a + b + c ^ 2}\\ \leq&\frac{1+b+c}{(a+b+c)^2}+\frac{1+a+c}{(x+y+z)^2}+\frac{1+x+y}{(x+y+z)^2}\\ =&\frac{3+2(a+b+c)}{(a+b+c)^2}\leq& 1\end{align*}Solution. The Cauchy-Schwarz's Inequality gives $(a+b+c)^2\le(a^2+b+c)(1+b+c)$ for any real numbers $a,b,c\ge0$, and thus \begin{align*}&{1 \over a ^ 2 + b+ c} + {1 \over a + b^ 2 + c} + {1 \over a + b + c ^ 2}\\ \leq&\frac{1+b+c}{(a+b+c)^2}+\frac{1+a+c}{(a+b+c)^2}+\frac{1+a+b}{(a+b+c)^2}\\ =&\frac{3+2(a+b+c)}{(a+b+c)^2}\leq&1\end{align*}