[toán 8] bpt nâng cao

[giang11820]

[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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1, $a^2 + b^2 + c^2 + d^2$ \geq a(b + c + d)
2, $\frac{a^2}{b^2}$ + $\frac{b^2}{c^2}$ + $\frac{c^2}{a^2}$ \geq $\frac{c}{b}$ + $\frac{b}{a}$ + $\frac{a}{c}$
3, (a+b+c)($\frac{1}{a}$ + $\frac{1}{b}$ + $\frac{1}{c}$) \geq 9

 

[hiendang241]

1/

ta có $a^2$+$b^2$+$c^2$+$d^2$\geq a(b+c+d)
\Leftrightarrow $a^2$+$b^2$+$c^2$+$d^2$\geq ab+ac+ad
\Leftrightarrow 4$a^2$+4$b^2$+4$c^2$+4$c^2$\geq 4ab+4ac+4ad
\Leftrightarrow $(a-2b)^2$+$(a-2c)^2$+$(a-2d)^2$+$a^2$\geq0(dpcm)​

 

[thaolovely1412]

3)áp dụng bất đẳng thức BUNHIAXCOPSKI với a , b ,c >0 ta được :
[TEX]9 = (sqrt{a}.\frac{1}{sqrt{a}} + sqrt{b}.\frac{1}{sqrt{b}}+sqrt{c}.\frac{1}{sqrt{c}})^2 \leq (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) [/TEX]
\Rightarrow [TEX](a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq 9[/TEX]
dấu = xảy ra \Leftrightarrow a=b=c

 

[nhuquynhdat]

2) $\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2} \ge 2\sqrt{\dfrac{a^2}{c^2}}=2.\dfrac{a}{c}$

$\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2} \ge 2\sqrt{\dfrac{c^2}{b^2}}=2.\dfrac{c}{b}$

$\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2} \ge 2\sqrt{\dfrac{b^2}{a^2}}=2.\dfrac{b}{a}$

$\to 2(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2})\ge 2(\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a})$

$\to \dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2} \ge \dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a}$

 

[soicon_boy_9x]

$1) a^2+b^2+c^2+d^2 \geq a(b+c+d)$

$ \leftrightarrow (\dfrac{a}{2}-b)^2+(\dfrac{a}{2}-b)^2+(\dfrac{a}{2}-b)^2 \geq 0$(BĐT luôn đúng)

$2)$ Áp dụng BĐT $a^2+b^2+c^2 \geq ab+bc+ca$

$3)(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} ) \geq 3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9$

 

[thaolovely1412]

2) Ta có: [TEX](x-y)^2 \geq 0 \forall x [/TEX]
\Leftrightarrow [TEX]x^2-2xy+y^2 \geq 0 \forall x[/TEX]
\Leftrightarrow [TEX]x^2+y^2 \geq 2xy[/TEX]
\Rightarrow [TEX]\frac{a^2}{b^2}+\frac{c^2}{a^2} \geq 2\frac{c}{b}[/TEX]
[TEX]\frac{b^2}{c^2}+\frac{c^2}{a^2} \geq 2\frac{b}{a}[/TEX]
[TEX]\frac{a^2}{b^2}+\frac{b^2}{c^2} \geq 2\frac{a}{c}[/TEX]
Do đó: [TEX]2( \frac{a^2}{b^2}+\frac{c^2}{a^2}+\frac{b^2}{c^2}) \geq 2(\frac{c}{b}+\frac{a}{c}+\frac{b}{a})[/TEX]
\Leftrightarrow [TEX]\frac{a^2}{b^2}+\frac{c^2}{a^2}+\frac{b^2}{c^2} \geq \frac{c}{b}+\frac{a}{c}+\frac{b}{a}[/TEX]

 

[hiendang241]

3/

ta có $(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})$
$=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+ \dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1$
$=3+(\dfrac{a}{b}+\dfrac{b}{a})+(\dfrac{a}{c}+ \dfrac{c}{a})+(\dfrac{c}{b}+\dfrac{b}{c})$

dễ dàng c/m đc dạng chung $\dfrac{x}{y}+\dfrac{y}{x}\ge 2$ vs x,y dương
$\rightarrow (a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})\ge 9$