Minimizing in 3 variables

Question

Find the least possible value of the fraction $\dfrac{a^2+b^2+c^2}{ab+bc}$, where $a,b,c > 0$.

My try:

$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,

$= (a+c)/b +b/(a+c) -2ac/b(a+c)$

AM > GM

$3\sqrt[3]{-2ac/b(a+c)}$

And I cant somehow move on.

Answer 1

You can do the following:

$$\frac{a^2+b^2+c^2}{b(a+c)}\geq \frac{\frac{1}{2}(a+c)^2+b^2}{b(a+c)}=\frac{1}{2}\frac{a+c}{b}+\frac{b}{a+c}$$

where at the first step we used $2(a^2+c^2)\geq (a+c)^2$

If you now set $\frac{a+c}{b}=x$ then you just have to minimize $\frac{x}{2}+\frac{1}{x}$ over the positive real numbers.

But you have that by AM-GM $\frac{x}{2}+\frac{1}{x}\geq \sqrt{2}$ with equality if $x=\sqrt{2}$, i.e. $a+c=\sqrt{2}b$

Going back even further we want $a=c$ from the first step

Answer 2

Write

$$\frac{2a^2+b^2+b^2+2c^2}{2}=\frac{2a^2+b^2}{2}+\frac{b^2+2c^2}{2}\geq \sqrt{2}ab+\sqrt{2}bc$$

Answer 3

Let $\dfrac{a^2+b^2+c^2}{ab+bc}=k>0$ as $a,b,c>0$

$\iff b^2-kb(a+c)+a^2+c^2=0$ which is a Quadratic Equation in $b$

As $b$ is real, the discriminant must be $\ge0$

i.e., $$k^2(a+c)^2-4(a^2+c^2)\ge0\iff k^2\ge\dfrac{4(a^2+c^2)}{(a+c)^2}$$

the equality occurs if $a=\dfrac{k(a+c)}2$

Now $2(a^2+c^2)-(a+c)^2=(a-c)^2\ge0\implies2(a^2+c^2)\ge(a+c)^2$

$\implies k^2\ge2$

the equality occurs if $a=c$

Answer 4

Yet another approach: The expression does not change if $(a, b, c)$ are multiplied by a common factor, so we can assume that $a+c=2$. Then $$ \frac{a^2+b^2+c^2}{ab+bc} = \frac 12 \left (\frac{a^2+(2-a)^2}{b} + b \right) \ge \sqrt{a^2 + (2-a)^2} = \sqrt{2(a-1)^2 + 2} \, , $$ using $AM \ge GM$. It follows that $$ \frac{a^2+b^2+c^2}{ab+bc} \ge \sqrt 2 \, , $$ with equality if and only if $(a, b, c)$ is a multiple of $(1, \sqrt 2, 1)$.