By scalling, we only need to prove that if one of them is equal to 1, the LHS is greater or equal to 1. This mean |1 + a + b| + |ab + b + a| + |ab| = |1 + a + b| + |-ab - a - b| + |ab| >= 1. The last part is a strait forward triangle inequality.

R8kt wrote: By scalling, we only need to prove that if one of them is equal to 1, the LHS is greater or equal to 1. This mean |1 + a + b| + |ab + b + a| + |ab| = |1 + a + b| + |-ab - a - b| + |ab| >= 1. The last part is a strait forward triangle inequality. How does scaling work, may I ask? Why do we only need to prove if one of them is equal to 1?

So assume the inequality holds and we replace (a, b, c) with (x*a, x*b, x*c) where |x| <= 1, the inequality is still.

amogususususus wrote: R8kt wrote: By scalling, we only need to prove that if one of them is equal to 1, the LHS is greater or equal to 1. This mean |1 + a + b| + |ab + b + a| + |ab| = |1 + a + b| + |-ab - a - b| + |ab| >= 1. The last part is a strait forward triangle inequality. How does scaling work, may I ask? Why do we only need to prove if one of them is equal to 1? you are right

R8kt wrote: By scalling, we only need to prove that if one of them is equal to 1, the LHS is greater or equal to 1. This mean |1 + a + b| + |ab + b + a| + |ab| = |1 + a + b| + |-ab - a - b| + |ab| >= 1. The last part is a strait forward triangle inequality. What is this?

I will provide a solution without scaling. Assume |x|> 1, then |x+y+z| + |xy + yz + zx| + |xyz| > |1 + y/x + z/x| + |y/z + z/x + (yz)/(x^2)| + |(yz)/(x^2)| = |1 + y/x + z/x| + |-y/z - z/x - (yz)/(x^2)| + |(yz)/(x^2)| >= 1 by triangle inequality.

R8kt wrote: I will provide a solution without scaling. Assume |x|> 1, then |x+y+z| + |xy + yz + zx| + |xyz| > |1 + y/x + z/x| + |y/z + z/x + (yz)/(x^2)| + |(yz)/(x^2)| = |1 + y/x + z/x| + |-y/z - z/x - (yz)/(x^2)| + |(yz)/(x^2)| >= 1 by triangle inequality. Nice solution .

amogususususus wrote: Given real numbers $x,y,z$ which satisfies $$|x+y+z|+|xy+yz+zx|+|xyz| \le 1$$Show that $max\{ |x|,|y|,|z|\} \le 1$. Let $z=a,$ $x+y=s$ and $xy=p.$ Assume by contradiction that $|a|>1.$ We have: $$1\geq|a+s|+|as+p|+|ap|\geq$$$$\geq|a+s|+\frac1{|a|}\cdot|as+p|+\frac1{a^2}\cdot|ap|=$$$$=|a+s|+\left|s+\frac pa\right|+\left|\frac pa\right|\geq$$$$\geq\left|a+s-s-\frac pa+\frac pa\right|=|a|>1.$$Contradiction! Pa!

I hope my solution is, of course, correct. Feedback, please

R8kt wrote: I will provide a solution without scaling. Assume |x|> 1, then |x+y+z| + |xy + yz + zx| + |xyz| > |1 + y/x + z/x| + |y/z + z/x + (yz)/(x^2)| + |(yz)/(x^2)| = |1 + y/x + z/x| + |-y/z - z/x - (yz)/(x^2)| + |(yz)/(x^2)| >= 1 by triangle inequality. Let me assist you with the LaTeX Assume $|x|> 1$, then \begin{align*} 1 &\ge |x+y+z| + |xy + yz + zx| + |xyz| \\ &> \frac{|x+y+z|}{|x|} + \frac{|xy+yz+zx|}{|x|^2}+ \frac{|xyz|}{|x|^3} \\ &= \left\lvert 1 + \frac{y}{x} + \frac{z}{x} \right\rvert + \left\lvert \frac{y}{x} + \frac{z}{x} + \frac{yz}{x^2} \right\rvert + \left\lvert \frac{yz}{x^2} \right\rvert \\ &= \left\lvert 1 + \frac{y}{x} + \frac{z}{x} \right\rvert + \left\lvert -\frac{y}{x} - \frac{z}{x} - \frac{yz}{x^2} \right\rvert + \left\lvert \frac{yz}{x^2} \right\rvert \\ &\ge \left\lvert 1 + \frac{y}{x} + \frac{z}{x} -\frac{y}{x} - \frac{z}{x} - \frac{yz}{x^2} + \frac{yz}{x^2} \right\rvert \\ &= 1 \end{align*}Contradiction, hence $|x|\le 1$. By symmetry, we have $|x|,|y|,|z|\le 1$. Hence $\text{max} \{|x|,|y|,|z|\}\le 1$. Proven

amogususususus wrote: Given real numbers $x,y,z$ which satisfies $$|x+y+z|+|xy+yz+zx|+|xyz| \le 1$$Show that $max\{ |x|,|y|,|z|\} \le 1$. My solution during the test: Assume otherwise there exist $x,y,z$ such that \[|x+y+z| + |xy+yz+zx| + |xyz| \leqslant 1\]and $\text{max}\{|x|,|y|,|z|\} > 1 $. By symmetry, WLOG $|x|>1$. Since triples $(x,y,z)$ and $(-x,-y,-z)$ are equivalent in this problem, WLOG $x\geqslant 0$. From $|x|>1$ and $x\geqslant 0$, we have $x>1$. We have $$1\geqslant |x+y+z| + |xy+yz+zx| + |xyz| \geqslant |x+y+z| \implies 1\geqslant x+y+z \implies y+z \leqslant 1-x < 0 $$Let $y+z = -a$ for $a>0$, let $yz=b$. Recall that $x>1$, $a>0$, $b\in \mathbb{R} $ that satisfies \[1 \geqslant |x-a|+|b-xa| + |bx| \] Claim : $|b-xa| + |bx| \geqslant xa$ Proof : We will divide into three cases Case I : $b \geqslant xa$, then $|-xa+b| + |xb| \geqslant |xb| \geqslant x^2a \geqslant xa $ Case II : $xa \geqslant b \geqslant 0$, then $|-xa+b| + |xb| = xa-b+xb = xa+(x-1)b \geqslant xa $ Case III : $0 \geqslant b $, then $|-xa+b| + |xb| = xa-b-xb = xa + (-b)(x+1) \geqslant xa $ In all cases, claim proven $\square$ Recall that $x>1, a>0$. Now we have \begin{align*} 1 &\geqslant |x-a|+|b-xa|+|bx| \\ &\geqslant |x-a|+xa \end{align*}Since $1 \geqslant |x-a|+xa \geqslant xa \implies 1\geqslant xa $. Since $x>1$, then $1>a$. So $x>1>a>0$. But now notice that \[1\geqslant |x-a|+xa = x-a+xa = x + a(x-1) \geqslant x > 1 \]Contradiction, so the assumption $\text{max}\{ |x|,|y|,|z| \} > 1 $ is false. $\therefore$ Hence it's proven for any reals $x,y,z$ that satisfies $|x+y+z| + |xy+yz+zx| + |xyz| \leqslant 1 $, it follows that $\text{max}\{|x|,|y|,|z|\} \leqslant 1 $.

Some caseworks also works. FTSOC, suppose $|x| > 1$. Then there are two cases: 1. $x > -1$. Then consider the following equation: $1+(x+y+z)+(xy+yz+xz)+(xyz)=(x+1)(y+1)(z+1)$. LHS is non-negative, thus also $y \leq -1$. Then because $|x+y+z| \leq 1$, $|z|>1$, so $|xyz|>1$ - contradiction. 2. $x > 1$. Then consider $(1-x)(1-y)(1-z)=1-(x+y+z)+(xy+yz+xz)-xyz$. Then RHS is once again non-negative, thus $y \geq 1$. Then because $|x+y+z| \leq 1$, $z \leq -1$, so $|xyz|>1$ - contradiction,