I am working through a pure maths book as a hobby and have come to inequalities. A question asks:
Prove that for any real numbers x,y,z $3(xy+yz+zx)\le(x+y+z)^2\le3(x^2+y^2+z^2)$
I expanded to get
$(x+y+z)^2=(x^2+y^2+z^2)+2(xy+xz+yz)$ but I cannot get from this to the answer. I am particularly puzzled how/why the 3 crops up.
On
$$(x+y+z)^2-3(xy+xz+yz)=\frac{1}{2}\sum_{cyc}(x-y)^2\geq0.$$ $$3(x^2+y^2+z^2)-(x+y+z)^2=\sum_{cyc}(x-y)^2\geq0.$$
On
If $A=xy+yz+zy$, $B=(x+y+z)^2$, $C=x^2+y^2+z^2$ then you have already shown: $$ B = C+2A $$
Now let’s proove: $A\leq C$. $A$ is a form given by $\langle X, MX\rangle$ where $X=(x,y,z)$ and $M(x,y,z)=(y,z,x)$. Then $$ A = \langle X, MX\rangle \leq ||X||\,||MX||= ||X||^2 = C$$
Thus $B-3A = C-A \geq 0$, which proves the first inequality. Also we get $B=C+2A\leq 3C$. This proves the second inequality.
On
Yes, go ahead with your method:
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+da)~~~~(1)$$ Due to AM-GM $a^2+b^2\ge 2ab$ etc, we know $$a^2+b^2+c^2 \ge ab+bc+ca~~~~(2)$$ Add (1) and (2) to get $$(a+b+c)^2 \ge 3(ab+bc+ca)$$ Re-write (2) as $$ab+bc+ca \le a^2+b^2+c^2 ~~~~~(3)$$ then add rwice of (2) to (1) to get $$(a+b+c)^2 \le 3(a^3+b^2+c^2)$$
This is a pretty standard form of rearrangement inequality. Without loss of generality, assume $$z\le y \le x$$ Then, $$z^2\le y^2\le x^2$$ Rearrangement inequality states, if two sequences,$$a_1,a_2...a_n$$ and $$b_1,b_2...b_n$$ are similarly sorted, either increasing or decreasing, then we have $$a_nb_n+...a_2b_2+a_1b_1\ge a_1b_n+...a_nb_1$$ Or in words, the maximum of the sum occurs when similarly sorted and the minimum when oppositely sorted. Applying rearrangment inequality on the set $x,y,z$ twice, we get $$xy+yz+zx\le x^2+y^2+z^2~~~~(1)$$ Adding $2xy+2yz+2zx$ to both sides,we get the first part of the inequality, $$3(xy+yz+zy)\le(x+y+z)^2$$ and since $$ (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\le x^2+y^2+z^2 +2(x^2+y^2+z^2) from (1)$$ $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\le 3(x^2+y^2+z^2)$$ we have proved the 2nd part of the inequality as well.