$f(x)=\frac{1}{1-x}$ olsun $f''(x)\>0$ konveks jensen kullanabiliriz $$f(a)+f(b)+f(c) \geq 3 f(\frac{a+b+c}{3})$$ buda $$\frac{1}{1-a^2}+\frac{1}{1-b^2}+\frac{1}{1-c^2}\geq 3 \frac{1}{1-\frac{a^2+b^2+c^2}{3}}=3.\frac{1}{1-\frac{1}{9}}=\frac{27}{8}$$
not: $a^2+b^2+c^2 \geq \frac{(a+b+c)^2}{3}=\frac{1}{3}$