# If $x^2 + y^2 + z^2 + 2xyz = 1$ then $x+ y+ z \leq \frac{3}{2}?$

1,154

If $z=0$ so $x+y+z=x+y\leq\sqrt{2(x^2+y^2)}=\sqrt2<\frac{3}{2}$.

Thus, it remains to prove our inequality for $xyz\neq0$.

Let $x=\frac{a}{\sqrt{(a+b)(a+c)}}$ and $y=\frac{b}{\sqrt{(a+b)(b+c)}}$,where $a$, $b$ and $c$ are positives.

Hence, $z=\frac{c}{\sqrt{(a+c)(b+c)}}$ and we need to prove that $\sum\limits_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}\leq\frac{3}{2}$, which is AM-GM: $$\sum\limits_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}=\sum_{cyc}\sqrt{\frac{a}{a+b}\cdot\frac{a}{a+c}}\leq$$ $$\leq\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+b}+\frac{b}{a+b}\right)=\frac{3}{2}.$$ Done!

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### medicu

Updated on August 01, 2022

• medicu 10 months

True or false? If $x \geq 0, y \geq 0, z \geq 0$ and $x^2 + y^2 + z^2 + 2xyz = 1$ then $x+ y+ z \leq \frac{3}{2}.$

I want to know if there is a way to demonstrate this conditional inequality. I know I can make a connection with two properties known in a triangle. I tried to find an algebraic demonstration of these problems and we did. Thanks in advance for any suggestions.

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