An algebra problem by Ayushmaan Seth

1 Let $a+b+c=1$ . The inequality reduces to $\sum_{cyc} \frac{1}{1-a}$ let $f(x)=\frac{x}{1-x}$ . Since $f(x)$ is a convex function, by Jensen's Inequality, we have $f(\frac{a+b+c}{3}) \leq \frac{f(a)+f(b)+f(c)}{3}$ so, we have $\sum_{cyc} \frac{1}{1-a} \geq 3 \times \frac{1}{2}=\boxed{1.5}$

Remember the cauchy shwartz ineqaulity.(Titu's lemma is a modified form) Mulitply both numerator and denominator by a in the first term, by b in second and by c in third. Now by titu's lemma- x^2/a + y^2/b + z^2/c >=( x+y+z) ^2 / a+b+c. Applying this in above we get minimum value 3/2(note- also used a^2 + b^2 + c^2 >= ab + bc +ca by AM-GM)