parmenides51 wrote: Find all the real numbers $a$ and $b$ that satisfy the relation $2(a^2 + 1)(b^2 + 1) = (a + 1)(b + 1)(ab + 1)$ Cauchy-Schwarz: $(a^2+1)(b^2+1) \geq \frac{1}{4}(a+1)^2(b+1)^2$ $(a^2+1)(b^2+1) \geq (ab+1)^2$ Sum and AM-GM: $2(a^2+1)(b^2+1) \geq \frac{1}{4}(a+1)^2(b+1)^2+(ab+1)^2 \geq (a+1)(b+1)(ab+1)$ Equality when $a=b=1$.

By Cauchy, we have: $\sqrt{(a^2+1)(b^2+1)} \geq ab+1$ and the equality holds only if $a=b$. So we have $(a+1)(b+1)(ab+1) = (ab+a+b+1)(ab+1) \leq (a^2+1)(b^2+1)+(a+b)\sqrt{(a^2+1)(b^2+1)}$ Substituting in the question equation: $2(a^2+1)(b^2+1)\leq (a^2+1)(b^2+1)+(a+b)\sqrt{(a^2+1)(b^2+1)}$ $\iff (a^2+1)(b^2+1)\leq(a+b)\sqrt{(a^2+1)(b^2+1)}$ $\iff \sqrt{(a^2+1)(b^2+1)} \leq a+b$ $\iff a^2b^2+a^2+b^2+1 \leq a^2 + 2ab + b^2 $ $\iff a^2b^2 -2ab + 1 \leq 0$ $\iff (ab-1)^2 \leq 0$ But $(ab-1)^2 \geq 0$ So $ (ab-1)^2 = 0 \iff ab = 1$ By the first inequality(Cauchy), we have that $a$ must be equal to $b$. So the only solution is $(a,b) = (1,1)$

Let $a, b$ be real numbers . Prove that$$2(1+a^2)(1+b^2)\geq(1+a)(1+b)(1+ab)$$

I'll provide a solution not using inequalities: $2(a^2+1)(b^2+1)=(a+1)(b+1)(ab+1)$ $\iff$ $2(a^2b^2+a^2+b^2+1)=(ab+a+b+1)(ab+1)$ $\iff$ $a^2b^2+2a^2+2b^2+2-2ab-a^2b-ab^2-a-b-1=0$, now the plan is to set a quadratic equation on $a$: $a^2(b^2-b+2)-a(b^2+2b+1)+2b^2-b+1=0$ and now using the well know: $\Delta=(b+1)^4-4(b^2-b+2)(2b^2-b+1)$ and after at little bit if algebra we get this surprise: $\Delta=-7b^4+16b^3-18b^2+16b-7$ and we realize that the sum of the coefficients is equal to $0$ so $b=1$ is a solution so by polynomial long division: $\Delta=(b-1)(-7b^3+9b^2-9b+7)$ and again the second term has $b=1$ being a solution $\implies$ $\Delta=(b-1)^2(-7b^2+2b-7)$ now we see that $\Delta=-7b^2+2b-7$ is negative for all $b$ but $(b-1)^2\geq0$ so $\Delta\le0$ and since $a$ is real then $\Delta=0$ therefore $b=1$ and using that in the beginning we'll have $a=1$ so $a=b=1$ is the only solution. $\square$.