Inequality with powers of 2

Question

If $m,n$ and $p $ are positive integers show that : $$\frac{2^{mn}}{p}+\frac{2^{np}}{m}+\frac{2^{pm}}{n}\geq 2(m+n+p).$$ I tried using Bernoulli inequality and then Hölder's, but I cannot prove this inequality.

Answer 1

Note that for $k\ge 1 \implies 2^k\geq 2k$ then

$$\frac{2^{mn}}{p}+\frac{2^{np}}{m}+\frac{2^{pm}}{n}\geq \frac{2{mn}}{p}+\frac{2{np}}{m}+\frac{2{pm}}{n}\geq 2(m+n+p)$$

which is true by Muirhead's inequality since

$$\frac{{mn}}{p}+\frac{{np}}{m}+\frac{{pm}}{n}\geq m+n+p \iff m^2n^2+n^2p^2+p^2n^2\geq m^2np+n^2mp+p^2mn$$

Answer 2

$$\frac{{mn}}{p}+\frac{{np}}{m}+\frac{{pm}}{n}\geq m+n+p$$ $$\frac{(mn)^2}{mnp}+\frac{(np)^2}{mnp}+\frac{(pm)^2}{mnp}\geq \frac{(mn+np+pm)^2}{3mnp}$$ $$\frac{(mn+np+pm)^2}{3mnp} \geq m+n+p$$ $$(mn-np)^2+(np-pm)^2+(pm-mn)^2 \geq 0$$