Inequality who involve squares

Question

Problem:

Let $xy+yz+xz = 26$ with $x,y,z \in\mathbb{Z}$. Show that $x^2+y^2+z^2 \geq 29$.

My try: It's well knowed the fact $(x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+xz) \geq 0$ so $x^2+y^2+z^2 \geq 52$. But, its not enough.

From other view i see $29 = 26 + 3$ so i tried using $AM-GM\text{ Ineq}$ but not work. Any ideas?

Answer 1

$$ x^2 + y^2 + z^2 - yz-zx-xy = \frac{1}{2} \left((y-z)^2 + (z-x)^2 +(x-y)^2 \right) \geq 0 $$ We get immediately $$ x^2 + y^2 + z^2 \geq yz+zx+xy = 26. $$

From the first expression, we see that equality holds only when $x=y=z.$

Now, restricting to integers: first, to get $x^2+y^2+z^2$ exactly equal to $26,$ we are required to have $x=y=z.$ This is not possible, since we would then have $3x^2=26$ for an integer $x.$

Finally $29=16+9+4$ for triple $(4,3,2)$ and we get $3 \cdot 2 + 2 \cdot 4 + 4 \cdot 3 = 6+8+12=26$

So far, $x^2 + y^2 + z^2 \geq 27.$

Next, there arer two ways to express $27$ as the sum of three squares. $27=9+9+9$ is not allowed, as then $yz+zx+xy$ is divisible by $3,$ while $26$ is not divisible by $3.$ We have another possibility, $27 = 25 + 1 + 1.$ This means the integer triple is $(x,y,z) = (\pm 5, \pm 1, \pm 1).$ This way, the sum $yz+zx+xy= \pm 5 \pm 5 \pm 1,$ so that $|yz+zx+xy| \leq 11.$

Finally, $28$ is not the sum of three integer squares. Try it!

But $29=16+9+4$ and $12 +8 +6 =26$

Answer 2

You can frame this as an optimization problem as below: $$\ \min \ x^2+y^2+z^2 \quad s.t. \ xy+yz+xz=26 $$ Writing Lagrangian we get, $$\ L(x, y, z, \mu)=x^2+y^2+z^2+\mu(xy+yz+xz-26) $$ Setting $\frac{\partial L}{\partial x}=0, \frac{\partial L}{\partial y}=0 \text{ and } \frac{\partial L}{\partial z}=0$ we get, $$\ 2x=-\mu(y+z), \ 2y=-\mu(x+z), \ 2z=-\mu(x+y) \\ \implies x=y=z=\sqrt{\frac{26}{3}} $$ Hence minimum value of $x^2+y^2+z^2$ is $26$.

I see this may not be helpful as domain is set of integers.

Answer 3

So we need to prove that $(x-y)^2+(y-z)^2+(z-x)^2 \geq 6$.

Note that $(x-y)^2+(y-z)^2+(z-x)^2=2((x-y)^2+(y-z)^2)+2(x-y)(y-z)$ is an even integer. So we prove that $(x-y)^2+(y-z)^2+(z-x)^2 \notin \{0,2,4\}$

• If $ (x-y)^2+(y-z)^2+(z-x)^2=0$ then $x=y=z$, then $3x^2=26$ never happen with s is integer.

• If $(x-y)^2+(y-z)^2+(z-x)^2=2$. There is only one case is that $2=0^2+1^2+1^2$, WOLG, we can assume that $(y-z)^2=0$ and $(x-y)^2=(z-x)^2=1$. So $y=z=x-1$ or $y=z=x+1$. So we get $3x^2 \pm 2x =26$, which has no integer solution.

• If $(x-y)^2+(y-z)^2+(z-x)^2=4$. There is only one case is that $4=0^2+0^2+2^2$, WOLG, we can assume that $(x-y)^2=(y-z)^2=0$, which follows $x=y=z$, then $3x^2=26$, no integer solution.

So $(x-y)^2+(y-z)^2+(z-x)^2\geq 6$.

Answer 4

The case $x=y=z$ is impossible, otherwise $3x^2=26,$ which is a contradiction.

Let $x=y$.

Thus, $$x^2+2xz=26$$ or $$x(x+2z)=26,$$ which is impossible again because $(x+2z)-x$ is an even number.

Id est, $x\neq y$, $x\neq z$ and $y\neq z$.

Let $x<y<z,$ $y=x+a$ and $z=a+b,$ where $a\geq1$ and $b\geq1$.

Thus, $$x^2+y^2+z^2=\sum_{cyc}(x^2-xy)+xy+xz+yz=\frac{1}{2}\sum_{cyc}(x-y)^2+26=$$ $$=\frac{1}{2}(a^2+b^2+(a+b)^2)+26\geq\frac{1}{2}(1+1+4)+26\geq29.$$