Prove that for a, b, c, d=0, $$\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d} \geq \frac{64}{a+b+c+d}$$
I can't produce the RHS whatever I do. Please help!

Question

Prove that for a, b, c, d>=0, $$\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d} \geq \frac{64}{a+b+c+d}$$
I can't produce the RHS whatever I do. Please help!

kinggeorge · Answer

I'm a little out of practice with this, but have you tried partial fraction decomposition?

blockcolder · Answer

You mean the RHS? But I can't factor a+b+c+d.

kinggeorge · Answer

That is a good point. I told you I was out practice with this. :(

kinggeorge · Answer

On a side note, I think a, b, c, d have to be strictly greater than 0.

kinggeorge · Answer

I think... What you might want to do is to suppose the relation holds, and then show that it holds if a, b, c, d are positive. 

To do this, subtract the RHS from both sides so you have an expression that's greater than 0. Then, just multiply things until you get a common denominator. This part is really nasty and I used Wolfram to help me here, but you get that $${w^2 x y+16 w^2 x z+4 w^2 y z+w x^2 y+16 w x^2 z+w x y^2-42 w x y z}$$$$+16 w x z^2+4 w y^2 z+4 w y z^2+x^2 y z+x y^2 z+x y z^2$$must be greater than 0. 

btw, x=a, y=b, z=c, w=d in the above expression.

kinggeorge · Answer

This seems like it might be easier to prove.

blockcolder · Answer

What about y and z?

kinggeorge · Answer

And actually, you can't assume without loss of generality. There's a few terms that would have to be in there that aren't to assume that.

kinggeorge · Answer

I keep making mistakes I don't notice. I really need to sleep. In the above expression, 

z=a, w=b, x=c, y=d

So in the worst case scenario, take y to be the greatest, and z to be the least. Then just stick x, and w in the middle somewhere.

kinggeorge · Answer

There's probably a better way to do this. In any case, I must sleep now. Good luck and good night.