Proving inequality using Lagrange multipliers.

Question

I started learing about Lagrange Multipliers and I got the following question:

Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}≥\frac{3}{2}$, for each $a,b,c>0$.

I'm not sure how to use Lagrange multipliers for it... any ideas?

Answer 1

Let $f(x,y,z) = \frac{x}{y+z} +\frac{y}{x+z} +\frac{z}{x+y} $ $$\begin{eqnarray*} \vec \nabla f \cdot \hat x &=& +\frac{1}{(y+z)} - \frac{y}{(x+z)^2}- \frac{z}{(x+y)^2} \\ \vec \nabla f \cdot \hat y &=& -\frac{x}{(y+z)^2} + \frac{y}{(x+z)}- \frac{z}{(x+y)^2} \\ \vec \nabla f \cdot \hat z &=& -\frac{x}{(y+z)^2} - \frac{y}{(x+z)^2}+ \frac{z}{(x+y)} \end{eqnarray*}$$

You should be able to demonstrate that $$\vec\nabla f=0 \implies x=y=z$$

and since $f(x,x,x)=\frac 32$ you are pretty well done.

Answer 2

This is Nesbitt's inequality. There are a number of proofs for that, e.g. here. I don't think that Lagrange Multipliers apply here.

Answer 3

Let the objective function be $$f = a(c+a)(a+b) + b(b+c)(a+b) + c(b+c)(c+a)$$ and the constraint be $$g = (a+b)(b+c)(c+a) - k.$$ Note that if $g=0$, then $$\frac{f}{k} = \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}.$$ Solving $\nabla f-\lambda g = 0$ and $g=0$ we find $f$ is minimized for $a=b=c=k^{1/3}/2$, with value $3k/2$. Thus, $f/k \ge 3/2$, which proves the inequality.

Answer 4

Since our inequality is homogeneous, we can assume $a+b+c=3$.

Also let $f(a,b,c,\lambda)=\sum\limits_{cyc}\frac{a}{3-a}+\lambda(a+b+c-3)-\frac{3}{2}$ $$\frac{\partial f}{\partial a}=\frac{3}{(3-a)^2}+\lambda$$.

Let in $(a,b,c)$ our $f$ gets a minimal value ($f$ is a continuous function and this thing happens on the compact).

Thus, in this point $\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0$, which gives $$\frac{3}{(3-a)^2}=\frac{3}{(3-b)^2}$$
and since $a+b<3$, we obtain: $a=b$, which says that $c=3-2a$ and it remains to prove that $$\frac{2a}{3-a}+\frac{3-2a}{2a}\geq\frac{3}{2},$$ where $0<a<\frac{3}{2}$, which gives $(a-1)^2\geq0$.

Also for $a\rightarrow0^+$... our inequality is obviously true.

Done!