How can they know what type of inequality to use?

Question

Let: $x,y,z \ge 0$ and $xy+yz+zx=1$. Prove that: $$\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \ge \frac{5}{2}$$

Here is the solution:

Square both sides and add $(xy+yz+zx)$ in LHS, we have: $$(xy+yz+zx)(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})^2 \ge \frac{25}{4}$$

which can be rewritten as:$$(\sum_{sym}^{}x^5y-\sum_{sym}^{}x^4y^2)+3(\sum_{sym}^{}x^5y-\sum_{sym}^{}x^3y^3)+(\sum_{sym}^{}x^4yz+14\sum_{sym}^{}x^3y^2z+38x^2y^2z^2)\ge 0$$ which is true by Murihead!

My questions are how can they knew how to square both sides and add $(xy+yz+zx)$ to $LHS$, and in the last line of solution, how can they knew when to use Murihead. I'm so confuse when to use Murihead, when to use Weighted AM-GM or Rearrangement inequality. Thank you for all your helps!

Answer 1

Alternative proof.

By squaring both side of the OP, we'll prove that$$(ab+bc+ca)\left(\sum_{cyc}\frac{1}{(b+c)^2}+2\sum_{cyc}\frac{1}{(b+a)(c+a)}\right)\ge \frac{25}{4}.\tag{*}$$ Notice that $$2(ab+bc+ca)\sum_{cyc}\frac{1}{(b+a)(c+a)}=4+\frac{4abc}{(a+b)(b+c)(c+a)}\ge 4. \tag{1}$$ WLOG, assuming that $c=\min\{a,b,c\}.$ We have $2(a+b)>(a+c)(b+c),$ thus \begin{align*} \sum_{cyc}\frac{1}{(b+c)^2}&=\frac{1}{(b+a)^2}+\frac{2}{(b+c)(a+c)}+\frac{(a-b)^2}{(b+c)^2(a+c)^2} \\&\ge \frac{1}{(b+a)^2}+\frac{2}{(b+c)(a+c)}+\frac{(a-b)^2}{4(a+b)^2}. \end{align*}

Proof for $2(a+b)>(a+c)(b+c).$ Since $c=\min\{a,b,c\},$ thus $1=ab+bc+ca\ge 3c^2$ and by AM-GM $$2(a+b)\ge a+c+b+c\ge 2\sqrt{(a+c)(b+c)}>(a+c)(b+c) \iff 4\ge (a+c)(b+c) \iff c^2<3.$$

Also,\begin{align*} \frac{1}{(b+a)^2}&=\frac{ab+bc+ca}{(b+a)^2}\\&=\frac{ab}{(b+a)^2}+\frac{c}{a+b}\\&=\frac{ab}{(b+a)^2}+\frac{2c^2}{(a+c)(b+c)}+\frac{c(c-a)(c-b)}{(a+b)(a+c)(b+c)}\\&\ge \frac{ab}{(b+a)^2}+\frac{2c^2}{(a+c)(b+c)}. \end{align*} Hence \begin{align*} (ab+bc+ca)\sum_{cyc}\frac{1}{(b+c)^2}&\ge \frac{ab}{(b+a)^2}+\frac{2c^2}{(a+c)(b+c)}+\frac{2(ab+bc+ca)}{(a+c)(b+c)}+\frac{(a-b)^2}{4(a+b)^2}\\&=\frac{9}{4}. \tag{2} \end{align*} From $(1)$ and $(2)$ we proved $(*).$

Hence, the proof is done. Equality holds at $a=b=1;c=0$ and its permutations.

Answer 2

It's just homogenization!

Muirhead works with homogeneous symmetric polynomials and homogenization helps to see a possibility to use Muirhead.

It's not always helps.

Another way.

Let $x+y+z\geq2.$

Thus, $$\sum_{cyc}\frac{1}{x+y}=\sum_{cyc}\frac{xy+xz+yz}{x+y}=x+y+z+\sum_{cyc}\frac{xy}{x+y}\geq$$ $$\geq x+y+z+\sum_{cyc}\frac{xy}{x+y+z}=x+y+z+\frac{1}{x+y+z}\geq\frac{5}{2}.$$

Let $x+y+z\leq2$.

Thus, by C-S $$\sum_{cyc}\frac{1}{x+y}=\frac{1}{x+y+z}\sum_{cyc}\frac{x+y+z}{x+y}=$$ $$=\frac{3}{x+y+z}+\frac{1}{x+y+z}\sum_{cyc}\frac{z}{x+y}\geq$$ $$\geq\frac{3}{x+y+z}+\frac{1}{x+y+z}\cdot\frac{(x+y+z)^2}{\sum\limits_{cyc}(xz+yz)}=$$ $$=\frac{3}{x+y+z}+\frac{x+y+z}{2}\geq\frac{5}{2}.$$

Answer 3

Hint.

Calling $f(x,y) = \frac{1}{x+y}$ we have

$$ F(x,y,z) = f(x,y)+f(y,z)+f(z,x) $$

and also

$$ \cases{ f(x,y) \le F(x,y,z)\\ f(y,z) \le F(x,y,z)\\ f(z,x) \le F(x,y,z)\\ } $$

so the solution is reduced to determine the minimum of the problems

$$ \cases{ \min_{x,y}f(x,y), \ \ \text{s.t.}\ \ \ x y =1, x\ge 0, y\ge 0\\ \min_{y,z}f(y,z), \ \ \text{s.t.}\ \ \ y z =1, y\ge 0, z\ge 0\\ \min_{z,x}f(z,x), \ \ \text{s.t.}\ \ \ z x =1, z\ge 0, x\ge 0\\ } $$