# Proving the AM-GM Inequality with Lagrange Multipliers

Exercise: Let $x_1,x_2,...,x_n$ be real positive numbers. Prove the arithmetic-geometric mean inequality, $(x_1x_2...x_n)^{1/n}\le (x_1+x_2+...+x_n)/n$.

Hint: Consider the function $f(x_1,x_2,...,x_n)=(x_1+x_2+...+x_n)/n$ subject to the constraint $x_1x_2...x_n=c$ a constant.

My work:

Consider $f(x_1,x_2,...x_n)=(x_1+x_2+...+x_n)/n$ on $S=\{(x_1,x_2,...,x_n) | x_1x_2...x_n=c\}$

$g(x_1,x_2,...,x_n)=x_1x_2...x_n-c=0$

$\nabla f(x_1,x_2,...,x_n)=(1/n,1/n...,1/n)$

$\nabla g(x_1,x_2,...,x_n)=(c/x_1,c/x_2,...c/x_n)$

$\nabla f = \lambda \nabla g$

$1/(n\lambda)=c/x_1=c/x_2=...=c/x_n$

$x_1=x_2=...=x_n=n\lambda c$

At the point $(n\lambda c, n\lambda c,... n\lambda c)$, $f$ on $S$ takes either a maximum or a minimum.

$f(n\lambda c, n\lambda c,...,n\lambda c)=n(n\lambda c)/n=n\lambda c$

$(n\lambda c*n\lambda c*...*n\lambda c)^{1/n}=((n\lambda c)^n)^{1/n}=n\lambda c = f(n\lambda c, n\lambda c,...,n\lambda c)$

This shows the equality case. To prove the inequality, I want to show that at the point $(n\lambda c,n\lambda c,...,n\lambda c)$, $f$ on $S$ takes a maximum minimum. I tried proving that the Hessian is negative positive definite, but I had trouble working with the second derivatives. I've looked up various proofs for the AM-GM inequality. The Lagrange ones didn't make full sense to me, so I thought I would turn to MSE. Any ideas? Thanks!

• $f$ has a minimum on $S$ in that point, not a maximum. To see that it has a minimum, you could for example note that $f$ attains arbitrarily large values on the hyperboloid $S$, is bounded below, and has only one critical point, which therefore must be a minimum. – Daniel Fischer Mar 4 '14 at 20:22
• Daniel, could you expand on your comment? Does f attain arbitrarily large values on S because c can be arbitrarily large? And why is f bounded below? – Jeff Mar 4 '14 at 21:26
• For any $K > 0$, we can consider the point $\xi(K) = (K,b,b,\dotsc,b)$, where $b = (c/K)^{1/(n-1)}$. Then $f(\xi(K)) > K/n$, thus $f$ attains arbitrarily large values on $S$. On the other hand we have $x_k > 0$ for $1\leqslant k \leqslant n$, so $f(x) > 0$. – Daniel Fischer Mar 4 '14 at 21:43
• See also Proofs of AM-GM inequality – punctured dusk Feb 20 '15 at 18:05

## 1 Answer

Note: It is more convenient for the calculations to assume (without loss of generality) that $c=1$.

The Hessian matrix of $f$ is equal to $$\left(\frac{d^2f(x)}{dx_idx_j}\right)_{i,j=1,\ldots n}=\mathbf{0}_{n\times n}$$ and thus positive semidefinite (the zero matrix). For example $$\frac{d^2f(x)}{dx_2dx_1}=\frac{d}{dx_2}\left(\frac{df(x)}{dx_1}\right)=\frac{d}{dx_2}\left(\frac{d}{dx_1}\left(x_1+x_2+\ldots+x_n\right)\frac{1}{n}\right)=\frac{d}{dx_2}\left(\frac{1}{n}\right)=0$$ Note that this is also immediate since $f$ is linear in $x_i$ and thus convex. However a linear function is also concave, which means that the Hessian matrix is also negative semidefinite. Combining the two you get that the Hessian matrix is the zero matrix. So this is not the way to prove that $f$ has a minimum in that point.

Since you know that at the point $x^*:=(n\lambda c,\ldots,n\lambda c)$ the value of $f$ is equal to $$f(x^*)=n\lambda c$$ and that at that point $f$ has either a maximum or a minimum, it is easier to find a point in $S$ that has a greater value, thus proving that $f$ has in $x^*$ a minimum. Firstly, determine that $$\lambda=\frac{c^{\frac{1}{n}-1}}{n}$$ so that in the proposed solution $x_i=c^{\frac{1}{n}}$ and $f\left(c^{\frac{1}{n}},\ldots,c^{\frac{1}{n}}\right)=\frac{nc^{\frac{1}{n}}}{n}=c^{\frac{1}{n}}.$ For $c=1$ you have that $x^*=(1,1,\ldots,1)$ and $f(x^*)=1.$ Now the admissible solution $$(x_1,x_2,\ldots,x_n)=\left(2,\frac{1}{2},1,\ldots,1\right)$$ yields a larger value for $f$, which is equal to $$f\left(2,\frac{1}{2},1,\ldots,1\right)=\frac{2+\frac{1}{2}+n-2}{n}=1+\frac{1}{2n}>1$$ implying that in the point $x^8$ the function $f$ has a minimum as desired.