Homogeneity in solving inequalities

Homogeneous inequalities are inequalities in which if you multiply all the variables by some positive real \(\lambda \), it is possible to simplify and go back to the original inequality. Such inequalities often allow to fix conditions on its variables without loss of generality. By combining the proprieties of homogeneity with some Classical Inequalities, it's often possible to find elegant shortcuts leading to elegant solutions.

Let \(x_{1},...x_{n}\) be variables and let \( A(x_{1},...x_{n})\) be an expression.

\(A(x_{1},...x_{n})\) is homogeneous of degree \(k\) if, for all \(\lambda > 0 \), we have the following : \(A(\lambda x_{1},...\lambda x_{n}) = \lambda^{k}A(x_{1},...x_{n}) \)

If we multiply the variables of \(x + y + z\) by \(\lambda > 0\) we get that \(\lambda x + \lambda y + \lambda z = \lambda (x+y+z)\). Therefore it's homogeneous of degree \(1\).

Also, \(xyz\) is homogeneous of degree \(3\) and \(xy + yz + zx\) is homogeneous of degree \(2\)

An inequality is said homogeneous if both of its sides have the same degree of homogeneity

\(\frac{x_{1}+...+x_{n}}{n} \geq \sqrt[n]{x_{1}x_{2}...x_{n}}\) is homogeneous.

The advantage of homogeneous inequalities is that we can multiply the variables by a factor without loss of generality. That's because if the (homogeneous) inequality is true for \(x_{1},...x_{n}\) then it is also true for \(\lambda x_{1},...\lambda x_{n}\).

We can assume, without loss of generality, that the variables verify an additional propriety such as \(x_{1}+...+x_{n}=1\) or \(x_{1}x_{2}...x_{n}=1\) or some other condition depending on the problem.

Suppose that we've demonstrated a homogeneous inequality for \(x_{1}+...+x_{n}=1\) beforehand. We can then argue that it is also true for \(x_{1}+...+x_{n}=k\) for any \(k\). Here is why :

If \(x_{1}+...+x_{n}=k\) then \(\frac{x_{1}+...+x_{n}}{k}=1\) which means that \(\frac{x_{1}}{k}+...+\frac{x_{n}}{k}=1\).

Now since we know that the inequality holds for \(\frac{x_{1}}{k},...\frac{x_{n}}{k}\) then we can simply multiply both sides by \(k\) and it won't cause any problem thanks to homogeneity !

But always be wary of \(k=0\), if allowed.

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