parmenides51 wrote: Let $a$ and $b$ be positive real numbers with $a^2 + b^2 =\frac12$. Prove that $$\frac{1}{1 - a}+\frac{1}{1-b}\ge 4.$$When does equality hold? (Walther Janous) Because $$\sum_{cyc}\left(\frac{1}{1-a}-2\right)=\sum_{cyc}\left(\frac{2a-1}{1-a}-(4a^2-1)\right)=\sum_{cyc}\frac{(2a-1)^2a}{1-a}\geq0.$$

Clearing the denominators and homogenizing, it suffices to show that $\sqrt{2a^2 + 2b^2} (3a + 3b) \ge 4a^2 + 4b^2 + 4ab$. Squaring and expanding, it reduces to showing that $(a - b)^2 (a^2 + 4ab + b^2) \ge 0$, which is obvious.

It is equivalent to $3(a+b) \ge 2+4ab$, which, upon letting $m=ab$, boils down verifying $\textstyle 3\sqrt{2m+\frac12}\ge 2+4m$. It is easily seen the solution set is $-1/8\le m\le 1/4$. As $1/2 = a^2+b^2\ge 2ab=2m$, we indeed have $1/4\ge m$, hence the conclusion.

arqady wrote: Because $$\sum_{cyc}\left(\frac{1}{1-a}-2\right)=\sum_{cyc}\left(\frac{2a-1}{1-a}-(4a^2-1)\right)=\sum_{cyc}\frac{(2a-1)^2a}{1-a}\geq0.$$ Very nice, arqady! I'm a fan of your quick elegant solutions! parmenides51 wrote: Let $a$ and $b$ be positive real numbers with $a^2 + b^2 =\frac12$. Prove that $$\frac{1}{1 - a}+\frac{1}{1-b}\ge 4.$$When does equality hold? (Walther Janous)

parmenides51 wrote: Let $a$ and $b$ be positive real numbers with $a^2 + b^2 =\frac12$. Prove that $$\frac{1}{1 - a}+\frac{1}{1-b}\ge 4.$$When does equality hold? (Walther Janous) You can use this method: https://artofproblemsolving.com/community/c6h2935046p26265746

grupyorum wrote: It is equivalent to $3(a+b) \ge 2+4ab$, which, upon letting $m=ab$, boils down verifying $\textstyle 3\sqrt{2m+\frac12}\ge 2+4m$. It is easily seen the solution set is $-1/8\le m\le 1/4$. As $1/2 = a^2+b^2\ge 2ab=2m$, we indeed have $1/4\ge m$, hence the conclusion. we can also prove that 3(a+b)-2-4ab>=0 by lagrange multipliers