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

Question

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.

Answer 1

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!

Answer 2

Let $(a,b,c)=(\lambda x,\lambda y,\lambda z)$, where $\lambda(x+y+z)=\frac32$. Then $a+b+c=\frac32$, and we have $$a^2+b^2+c^2+2abc=\lambda^2(x^2+y^2+z^2+2\lambda xyz),$$ which is increasing in $\lambda$. Hence to show $\lambda\geq1$ it suffices to prove $$a+b+c=\frac32\implies a^2+b^2+c^2+2abc\geq1.$$ We now use the notation for cyclic sums $\sum$ and symmetric sums $\sum_\text{sym}$. Note that $(\sum a)^2=\sum a^2+2\sum ab=\frac94$, so \begin{align*} \sum a^2+2\sum abc&\geq1\\ \iff\frac94\left(\sum a^2+2\sum abc\right)&\geq\sum a^2+2\sum ab\\ \iff18abc&\geq8\sum ab-5\sum a^2\\ \iff27abc&\geq\left(\sum a\right)\left(8\sum ab-5\sum a^2\right)\\ &=8\sum_\text{sym}a^2b+24abc-5\sum a^3-5\sum_\text{sym}a^2b\\ \iff5\sum a^3+3abc&\geq3\sum_\text{sym}a^2b. \end{align*} The last inequality follows from AM-GM ($\sum a^3\geq3abc$) and Schur's inequality ($\sum a^3+3abc\geq3\sum_\text{sym}a^2b$): $$5\sum a^3+3abc\geq3\left(\sum a^3+3abc\right)\geq3\sum_\text{sym}a^2b,$$ and we are done.