Emirhan wrote: Let $a$,$b$,$c$ and $d$ positive reals. Prove that $$\frac{1}{a+b+c+d} \leq \frac{1}{64}(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d})$$ $\text{Cauchy-Schwarz inequality}$ implies $\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\geq \frac{(1+1+2+4)^2}{a+b+c+d}=\frac{64}{a+b+c+d},$ so $\frac{1}{a+b+c+d} \leq \frac{1}{64}\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\right).$

Titu's lemma kills it

Let $a,b,c,d\in(0,1]$ . Prove that $$\frac{1}{a+b+c+d}\geq\frac{1}{4}+\frac{64}{27}(1-a)(1-b)(1-c)(1-d)$$

sqing wrote: Let $a,b,c,d\in(0,1]$ . Prove that $$\frac{1}{a+b+c+d}\geq\frac{1}{4}+\frac{64}{27}(1-a)(1-b)(1-c)(1-d)$$ By AM-GM $(1-a)(1-b)(1-c)(1-d)\leq\left(\frac{4-a-b-c-d}{4}\right)^4$. Let $a+b+c+d=t$. Hence, $0<t\leq4$ and we need to prove that $\frac{1}{t}\geq\frac{1}{4}+\frac{(4-t)^4}{108}$ which is $(4-t)(t-1)^2(t^2-10t+27)\geq0$. Done!

sqing wrote: Let $a,b,c,d\in(0,1]$ . Prove that $$\frac{1}{a+b+c+d}\geq\frac{1}{4}+\frac{64}{27}(1-a)(1-b)(1-c)(1-d)$$ Yes, thank you. You are great.

Let $a,b,c,d\in(0,1]$ . Prove that $$\frac{1}{a+b+c+2d}\geq\frac{1}{5}+\frac{131072}{84375}(1-a)(1-b)(1-c)(1-d)$$