Advance Happy New Year 2015!

The expression can be split and rewritten as:

$\frac{x^2}{z} + \frac{y^2}{z} + \frac{y^2}{x} + \frac{z^2}{x} + \frac{z^2}{y} + \frac{x^2}{y}$

Applying Titu's Lemma Theorem to the above expression, we get:

$\frac{x^2}{z} + \frac{y^2}{z} + \frac{y^2}{x} + \frac{z^2}{x} + \frac{z^2}{y} + \frac{x^2}{y} \geq \frac{4(x + y + z)}{2(x + y + z)}$

Solving the above inequality and substituting the value of $x + y + z$ , we get:

$\frac{x^2 + y^2}{z} + \frac{z^2 + x^2}{y} + \frac{y^2 + z^2}{x} \geq 4030$

Thus, the minimum value of the expression is $4030$

$\frac { { x }^{ 2 }+{ y }^{ 2 } }{ z } +\frac { { y }^{ 2 }+{ z }^{ 2 } }{ x } +\frac { { z }^{ 2 }+{ x }^{ 2 } }{ y } =\frac { xy({ x }^{ 2 }+{ y }^{ 2 })+yz({ y }^{ 2 }+{ z }^{ 2 })+xz({ z }^{ 2 }+{ x }^{ 2 }) }{ xyz } \\ Now,\\ \qquad \qquad xy({ x }^{ 2 }+{ y }^{ 2 })\quad \ge \quad xy(2xy)\quad =\quad 2{ x }^{ 2 }{ y }^{ 2 }\\ So,\\ \qquad \frac { { x }^{ 2 }+{ y }^{ 2 } }{ z } +\frac { { y }^{ 2 }+{ z }^{ 2 } }{ x } +\frac { { z }^{ 2 }+{ x }^{ 2 } }{ y } \quad \ge \quad \frac { 2{ x }^{ 2 }{ y }^{ 2 }+2{ y }^{ 2 }{ z }^{ 2 }+2{ z }^{ 2 }{ x }^{ 2 } }{ xyz } \\ Now,\\ \qquad \qquad 2{ x }^{ 2 }{ y }^{ 2 }+2{ y }^{ 2 }{ z }^{ 2 }+2{ z }^{ 2 }{ x }^{ 2 }=({ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 2 }{ z }^{ 2 })+({ z }^{ 2 }{ x }^{ 2 }+{ x }^{ 2 }{ y }^{ 2 })+({ y }^{ 2 }{ z }^{ 2 }+{ z }^{ 2 }{ x }^{ 2 })\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ge 2({ x }^{ 2 }yz+x{ y }^{ 2 }z+xy{ z }^{ 2 })\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad =2xyz(x+y+z)\\ So,\\ \qquad \frac { { x }^{ 2 }+{ y }^{ 2 } }{ z } +\frac { { y }^{ 2 }+{ z }^{ 2 } }{ x } +\frac { { z }^{ 2 }+{ x }^{ 2 } }{ y } \quad \ge \quad \frac { 2xyz(x+y+z) }{ xyz } \quad =\quad 2(x+y+z)\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad =4030$

Since this equation is symmetric about $x = y = z$ , I guessed (correctly, apparently), that the mimimum value does in fact lie on $x = y = z$ . Plugging in $\frac{2015}{3}$ for x, y, and z, the equation simplifies to $2 * 2015 = \boxed{4030}$ .

It was an extremely lucky guess, though, as this equation could easily have had three (or even six) separate minimum points which evaulate to the same value. That sort of outcome would be more likely for a trig-type equation, though, as those tend to oscillate around the origin in very strange patterns.

I tried solving this before using Lagrange Multipliers, but the equations tried to eat me.

First, it's trivial to prove that $2({ a }^{ 2 }+{ b }^{ 2 })\ge { (a+b) }^{ 2 }$ .

Applying this, we get that $\frac { { x }^{ 2 }+{ y }^{ 2 } }{ z } =\frac { 2({ x }^{ 2 }+{ y }^{ 2 }) }{ 2z } \ge \frac { { (x+y) }^{ 2 } }{ 2z }$

$\frac { { y }^{ 2 }+{ z }^{ 2 } }{ x } =\frac { 2({ y }^{ 2 }+{ z }^{ 2 }) }{ 2x } \ge \frac { { (y+z) }^{ 2 } }{ 2x }$

$\frac { { z }^{ 2 }+{ x }^{ 2 } }{ y } =\frac { 2({ z }^{ 2 }+{ x }^{ 2 }) }{ 2y } \ge \frac { { (z+x) }^{ 2 } }{ 2y }$ .

Also, applying T2's lemma (Titu's lemma), we have that:

$\frac { { (x+y) }^{ 2 } }{ 2z } + \frac { { (y+z) }^{ 2 } }{ 2x } + \frac { { (z+x) }^{ 2 } }{ 2y } \ge \frac { { (x+y+y+z+z+x) }^{ 2 } }{ 2x+2y+2z } =\frac { [2{ (x+y+z)] }^{ 2 } }{ 2(x+y+z) } =2(x+y+z)$

Hence, $\frac { { x }^{ 2 }+{ y }^{ 2 } }{ z } + \frac { { y }^{ 2 }+{ z }^{ 2 } }{ x } +\frac { { z }^{ 2 }+{ x }^{ 2 } }{ y } \ge \frac { { (x+y) }^{ 2 } }{ 2z } + \frac { { (y+z) }^{ 2 } }{ 2x } + \frac { { (z+x) }^{ 2 } }{ 2y } \ge 2(x+y+z) = \boxed{4030}$ .

if x+y+z=2015 it should be divided equally. so that, the value of x=2015/3 then we a x=y=z

therefore : (x²+y²)/z + (z²+x²)/y + (y²+z²)/x =[(2015/3)² + (2015/3)²] / (2015/3)=1343.333333 then ; 1343.333333+1343.333333+1343.333333=4030 and (1343.333333)x(3)=4030

We can apply titu's lemma and solve this question - We can rewrite it and simply use the lemma

It's as easy as realizing the value of x, y, and z and then plugging the numbers in and simplifying. There is no need for the large expression until after you know the numbers.

Because you are looking for the minimum value of the expression, you want x, y, and z to be as large as possible since they will be the denominators (the larger the denominator of a fraction the smaller the value will reduce to).

I simply divided 2015 by 3 , as that will allow for the largest values of each of the 3 variables. So x, y, and z all equal 2015/3.

Next, plug 2015/3 in for all of the variables in the expression and simplify. You get 1343.3333... for each fraction individually and there for 4030 for all 3 added together.

So the answer to the problem is 4030.

The expression may be rewritten as S=x^2/z+x^2/y+... and that sum cyclic. Applyin CS, you obtain S=>{2(x+y+z)=4030