# Reciprocal inequality

Given that $$a, b, c, d$$ are greater than zero, prove that $$\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} ≥ \frac{64}{a+b+c+d}$$ The first thing that came to mind was QM-AM-GM-HM, but I haven't been able to create the second term with HM-AM since the geometric mean or the quadratic mean don't really make sense. $$\frac{4}{a+b+\frac{c}{4}+\frac{d}{16}} = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$$

I could probably multiply both sides by the denominator, but that seems too hard for a contest-like question. How can I solve this? Any help is appreciated.

• Cauchy-Schwarz. Jan 29 at 21:56
• Multiplying both sides by the denominator works, and it's just AM-GM on specific terms then (though be careful with the equality case). Give it a try. Jan 29 at 22:38

$$\frac{1^2}{a}+\frac{1^2}{b}+\frac{2^2}{c}+\frac{4^2}{d}\ge\frac{(1+1+2+4)^2}{a+b+c+d}$$
Equality holds for $$\frac{1}{a}=\frac{1}{b}=\frac{2}{c}=\frac{4}{d}=k$$ for some constant $$k$$.