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ZETA_in_olympiad wrote: If positive real numbers $x,y,z$ satisfies $x+y+z=3,$ prove that $$\sum_{\text{cyc}} y^2z^2<3+\sum_{\text{cyc}} yz.$$ Proposed by Li4 and Untro368. We'll extend to nonnegative numbers. Fix $xy+yz+zx=3(1-t^2);~0\leq t\leq1.$ It's well known that for $0\leq t\leq\frac12,$ $\max\left(\sum_{\text{cyc}} y^2z^2\right)=3(1+t)^2(3t^2-2t+1)$ and for $\frac12\leq t\leq1,$ $\max\left(\sum_{\text{cyc}} y^2z^2\right)=9(1-t^2)^2.$ In the first case we need to prove $$f(t):=(1+t)^2(3t^2-2t+1)+t^2-2<0,$$which is obviously true since $f$ is strictly increasing and $f\left(\frac12\right)<0.$ In the second case we need to prove $3a^2-a-1<0,$ for $1-t^2:=a\in\left[0,\frac34\right].$ But this is obvious

Remark: From my above proof, we deduce the following (optimal) inequality: If positive real numbers $x,y,z$ satisfy $x+y+z=3,$ then $$\sum_{\text{cyc}} y^2z^2<\frac{45}{16}+\sum_{\text{cyc}} yz.$$

W.L.O.G $x\leq y\leq z$. Because $x+y+z=3$ we have $3\geq xy+yz+xz$ which follows that $xy\leq 1$ and $x\leq 1$. We solve the problem in three cases: Case 1: $x\leq y\leq 1 \leq z$. Then $x^2 y^2+y^2z^2+x^2 z^2\leq xy+z^2(x+y)$. So we have $$xy+z^2(x+y)< 3+xy+yz + xz \leftrightarrow (x+y)(z^2-z)<3 \leftrightarrow (3-z)(z^2-z)<3 \leftrightarrow z(z-2)^2+(3-z)>0.$$ Case 2: $x\leq 1\leq y\leq z$ and $xz\leq 1$. Then $x^2 y^2+y^2z^2+x^2 z^2\leq xz + xy + y^2z^2 $. So $$xz + xy + y^2z^2 < 3+xy+xz+yz \leftrightarrow y^2z^2 -yz-3<0.$$Let $t = yz$. We know that $t\leq \dfrac{(y+z)^2}{4} \leq \dfrac{9}{4}$ that gives us $t^2-t-3<0$. Case 3. $x\leq 1\leq y\leq z$ and $xz\geq 1$. By $3\geq xy+yz+xz$, in this case we get $xz,\; yz\leq 2$. So $$x^2 y^2+y^2z^2+x^2 z^2\leq xy+2yz + 2xz \leq 3+ yz+xz < 3+xy+yz+xz.$$

mihaig wrote: Remark: From my above proof, we deduce the following (optimal) inequality: If positive real numbers $x,y,z$ satisfy $x+y+z=3,$ then $$\sum_{\text{cyc}} y^2z^2<\frac{45}{16}+\sum_{\text{cyc}} yz.$$ Le bump