## inequality – When can we deduce \$(mA+nB+pC)>(mX+nY+pZ)\$ from \$(A+B+C)>(X+Y+Z)\$?

I am trying to prove an equality in the form $$(mA+nB+pC)>(mX+nY+pZ)$$.
I can prove the inequality $$(A+B+C)>(X+Y+Z)$$ and I wonder if there is any condition,
under which we can deduce $$(mA+nB+pC)>(mX+nY+pZ)$$ from $$(A+B+C)>(X+Y+Z)$$ ?

And $$A,B,C,X,Y,Z,m,n,p$$ are all positive.

The first thing comes to mind is the Cauchy-Schwarz Inequality:

$$sqrt{left(m^{2}+n^{2}+p^{2}right)left(X^{2}+Y^{2}+Z^{2}right)}geleft(mX+nY+pZright)$$

Now it suffices to prove that

$$left(mA+nB+pCright)gesqrt{left(m^{2}+n^{2}+p^{2}right)left(X^{2}+Y^{2}+Z^{2}right)}$$
$$iffleft(mA+nB+pCright)^{2}geleft(m^{2}+n^{2}+p^{2}right)left(X^{2}+Y^{2}+Z^{2}right)$$

But I can’t find the condition under which we can deduce the above inequality from $$(A+B+C)>(X+Y+Z)$$.

Or using Hölder’s inequality:

$$left(mA+nB+pCright)^{frac{1}{2}}left(frac{1}{m}+frac{1}{n}+frac{1}{p}right)^{frac{1}{2}}ge A^{frac{1}{2}}+B^{frac{1}{2}}+C^{frac{1}{2}}$$
$$iffleft(mA+nB+pCright)^{frac{1}{2}}geleft(A^{frac{1}{2}}+B^{frac{1}{2}}+C^{frac{1}{2}}right)left(frac{1}{m}+frac{1}{n}+frac{1}{p}right)^{-frac{1}{2}}$$

And so on.