crazyfehmy wrote: Prove that \[ a^2b^2(a^2+b^2-2) \geq (a+b)(ab-1) \] for all positive real numbers $a$ and $b.$ Let $a+b=2u$ and $ab=v^2$, where $v>0$. Hence, $a^2b^2(a^2+b^2-2) \geq (a+b)(ab-1)\Leftrightarrow$ $\Leftrightarrow v^4(4u^2-2v^2-2)\geq2u(v^2-1)\Leftrightarrow2v^4u^2-(v^2-1)u-v^4(v^2+1)\geq0$. Id est, for the proof we need to prove that $u\geq\frac{v^2-1+\sqrt{(v^2-1)^2+8v^8(v^2+1)}}{4v^4}$. But $u\geq v$. Thus, it remains to prove that $v\geq\frac{v^2-1+\sqrt{(v^2-1)^2+8v^8(v^2+1)}}{4v^4}$, which is $(v-1)^2(v+1)(v^2+v+1)\geq0$. Done!

$a^2b^2(a^2+b^2-2)-(a+b)(ab-1)$ $\geq a^2b^2(2ab-2)-(a+b)(ab-1)\ \because a^2+b^2\geq 2ab$ $=(1-ab)(a+b-2a^2b^2)$ $\geq (1-ab)\{2\sqrt{ab}-2(\sqrt{ab})^4\}\ \because a+b\geq 2\sqrt{ab}$ $=2\sqrt{ab}(1+\sqrt{ab})(1+\sqrt{ab}+ab)(1-\sqrt{ab})^2\geq 0\ Q.E.D.$

kunny wrote: $(1-ab)(a+b-2a^2b^2) \geq (1-ab)\{2\sqrt{ab}-2(\sqrt{ab})^4\}\ \because a+b\geq 2\sqrt{ab}$ This is true only if $ab \leq 1.$

Equivalent to: $a^4 b^2 +a^2 b^4 +a+b \ge 2a^2 b^2 +a^2 b+ab^2$ proved by adding 3 ineuqualities by (weighted) AM-GM $\frac{1}{5} (2a^4 b^2 +2a+b) \ge a^2 b$ $\frac{1}{5} (2a^2 b^4 +a+2b) \ge ab^2$ $\frac{1}{5} (3a^4 b^2 +3a^2 b^4 +2a+2b) \ge 2a^2 b^2$.

crazyfehmy wrote: Prove that \[ a^2b^2(a^2+b^2-2) \geq (a+b)(ab-1) \] for all positive real numbers $a$ and $b.$ Let $a+b=2x$ and $\sqrt{ab}=y$. Then we have $x\ge y$ and inequality is equivalent to: $(2y^4)x^2+(1-y^2)x-y^4(y^2+1) \ge 0$ Define $f(x)=(2y^4)x^2+(1-y^2)x-y^4(y^2+1)$, then $f(x)$ reaches minimum at $x_0=\frac{y^2-1}{4y^4}$, but since $\frac{y^2-1}{4y^4}<y$ for all $y\ge 0$, it is enough to prove $f(x) \ge 0$ for $x=y$: $\Leftrightarrow y(y^5-y^3-y^2+1) \ge 0$ But this follows from rearrangement inequality: $y^5+1 \ge y^3+y^2$

prove of Kunny isn't right，if ab＞1 ？

crazyfehmy wrote: kunny wrote: $(1-ab)(a+b-2a^2b^2) \geq (1-ab)\{2\sqrt{ab}-2(\sqrt{ab})^4\}\ \because a+b\geq 2\sqrt{ab}$ This is true only if $ab \leq 1.$ No, by AM-GM we have $\dfrac{a+b}{2} \ge \sqrt{ab} \Longrightarrow a+b \ge 2\sqrt{ab}$ for all $a,b \ge 0$.

chaotic_iak wrote: crazyfehmy wrote: kunny wrote: $(1-ab)(a+b-2a^2b^2) \geq (1-ab)\{2\sqrt{ab}-2(\sqrt{ab})^4\}\ \because a+b\geq 2\sqrt{ab}$ This is true only if $ab \leq 1.$ No, by AM-GM we have $\dfrac{a+b}{2} \ge \sqrt{ab} \Longrightarrow a+b \ge 2\sqrt{ab}$ for all $a,b \ge 0$. But if $ab>1,$ then you cannot say that $(1-ab)x \geq (1-ab)y$ if $x \geq y.$ Is it okay now?