If $x^2 + y^2 + z^2 + 2xyz = 1$ then $x+ y+ z \leq \frac{3}{2}?$
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 AMGM: $$\sum\limits_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}=\sum_{cyc}\sqrt{\frac{a}{a+b}\cdot\frac{a}{a+c}}\leq$$ $$\leq\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+b}+\frac{b}{a+b}\right)=\frac{3}{2}.$$ Done!
Related videos on Youtube
medicu
Updated on August 01, 2022Comments

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.

lab bhattacharjee about 6 years

lab bhattacharjee about 6 years

medicu about 6 yearslab bhattacharjee@: I already said this: "I know I can make a connection with two properties known in a triangle".


medicu about 6 yearsMichael Rozenberg@: I expected you to use just wonderful substitutions. Thank you!