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

I am trying to prove an inequality 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:

$$\left(m^{2}+n^{2}+p^{2}\right)\left(X^{2}+Y^{2}+Z^{2}\right)\ge{\left(mX+nY+pZ\right)}^2$$

Now it suffices to prove that

$$\iff\left(mA+nB+pC\right)^{2}\ge\left(m^{2}+n^{2}+p^{2}\right)\left(X^{2}+Y^{2}+Z^{2}\right)$$

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

Another try:

$$\left(mA+nB+pC\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge( A^{\frac{1}{2}}+B^{\frac{1}{2}}+C^{\frac{1}{2}})^2$$

Now it suffices to show that

$$\left(A^{\frac{1}{2}}+B^{\frac{1}{2}}+C^{\frac{1}{2}}\right)^{2}>\left(mX+nY+pZ\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)$$

or

$$2\left(\sqrt{AB}+\sqrt{BC}+\sqrt{CA}\right)\ge mX\left(\frac{1}{n}+\frac{1}{p}\right)+nY\left(\frac{1}{p}+\frac{1}{m}\right)+pZ\left(\frac{1}{m}+\frac{1}{n}\right)$$

I wonder if there is a rule or something except $$A>X,\ B>Y$$, and $$C>Z$$.

• If the condition $A\ge X,B\ge Y,C\ge Z$ does not hold the aim inequality always can be broken by a suitable choice of $m,n,p$. – user Feb 23 at 16:08