Proving inequalities using linear function
Theorem1 ให้ $f$ เป็นฟังก์ชันซึ่ง $f(x)=ax+b$ ซึ่ง $f(\alpha) \ge 0$ และ $f(\beta) \ge 0$
แล้ว $f(x) \geq 0,\forall x \in [\alpha,\beta]$
Ex1 กำหนดให้ $x,y,z \in R^+$ และ $x+y+z=3$ จงพิสูจน์ว่า $x^2+y^2+z^2+xyz \ge 4$
Solution by Pham Van Thuan,Trieu Van Hung We rewrite the desired inequality in the form
$(y+z)^2-2yz+x^2+xyz \ge 4$
or $yz(x-2)+2x^2-6x+5 \ge 0$
Set $yz=w$,and view the expression on the left hand as a linear function of $w$,that is
$f(w)=(x-2)w+2x^2-6x+5$
Now we need to find all possible values of $w$. By AM-GM inequality,
$yz \leq \frac{(y+z)^2}{4}$.that is $w \leq \frac{(3-x)^2}{4}$,we also have
$w \ge 0$ by hypothesis. By the theorem 1,it's sufficient to show that $f(0) \ge 0$ and $f(w_0) \ge 0$.where $w_0=\frac{(3-x)^2}{4}$.It's easy to check that $f(0)=2x^2-6x+5=2(x-\frac{3}{2})^2+\frac{1}{5} \ge 0,$
$f(w_0)=\frac{1}{4}(x-1)^2(x+2) \ge 0$
The proof is complete.The equality holds if and only if all the three numbers are
equal to 1.
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