Cool inequality

Algebra Level 5

Let $x, y, z$ be positive real numbers such that: $x^2+y^2+z^2+1=2(xy+yz+zx)$ .

The minimum value of $P=3x+2y+6z$ is $\alpha$ iff $x=\beta; y=\gamma; z=\delta$ .

And $\alpha+\beta+\gamma+\delta=\dfrac{m}{n}$ where $m$ and $n$ are co-prime positive integers.

What is the value of $m+n$ ?

1 solution

Khang Nguyen Thanh
Aug 1, 2015

Let: $a=-x+y+z; b=z+x-y; c=x+y-z$ , we get: $x=\dfrac{b+c}{2}; y=\dfrac{c+a}{2}; z=\dfrac{a+b}{2}$

From the hypothesis, we have: $ab+bc+ca=1$ .

And $2P=2(3x+2y+6z)=8a+9b+5c$ .

Applying the AM-GM inequality, we get:

$(8a+9b+5c)^2=64a^2+81b^2+25c^2+144ab+90bc+80ca\\ =(64a^2+16c^2)+(81b^2+9c^2)+144ab+90bc+80ca\\ \ge 64ac+54bc+144ab+90bc+80ca\\ =144$

This implies that $8a+9b+5c\ge12$ or $3x+2y+6z\ge6$

The equality holds iff: $x=\dfrac{2}{3}; y=\dfrac{3}{4}; z=\dfrac{5}{12}$ .

We have: $\alpha=6; \beta=\dfrac{2}{3}; \gamma=\dfrac{3}{4}; \delta=\dfrac{5}{12}$ and $\alpha+\beta+\gamma+\delta=\dfrac{47}{6}$

So, we get $m=47, n=6$ and $m+n=\boxed{53}$

Nice method.However, I think Lagrange multipliers kills this problem.

Tobi Major - 5 years, 8 months ago

No, Using Lagrangian multipliers will be bit lengthy calculation but very easy. By the way, Nice Method.

Surya Prakash - 5 years, 8 months ago

I'm not familiar with the very first substitution of $a=-x+y+z,b=z+x-y, c=x+y-z$ . May I know what technique is this?

Pi Han Goh - 5 years, 6 months ago