Chủ đề này có 3 trả lời

[quanrrom97]

• $\frac{a^{8}+b^{8}+c^{8}}{a^{3}b^{3}c^{3}}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ (a,b,c>0)
• $\sum \frac{a}{2a+b+c}\leq \frac{3}{4}$ (a,b,c>0)
• $\frac{9}{a+b+c}\leq \sum \frac{4}{2a+b+c}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ (a,b,c>0)
• $\sqrt{x^{2}+x+1}+\sqrt{x^{2}-x+1}\geq 2$
• $\sum \sqrt{\frac{a+b}{c}}\geq \sum 2\sqrt {\frac{c}{a+b}}$ (a,b,c>0)
• $\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\geq 1$ (a,b,c>0)
• $\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\geq \frac{2}{3}$ (a,b,c,d>0)
• CM nếu $\left | b \right |<\frac{\left | a \right |}{2}$ thì $\frac{1}{\left | a-b \right |} < \frac{2}{\left | a \right |}$
• CM nếu $\left | a \right |\leq 2$ thì $x^{2}+axy+y^{2}\geq 0$ ( với mọi x,y)

[lovemoon]

$\frac{a}{b+2c+3d}\dotplus \frac{b}{c+2d+3a}\dotplus \frac{c}{d+2a+3b}\dotplus \frac{d}{a+2b+3c}\geq \frac{2}{3}$
đây là đề thi IMO shortlist 1993
$VT= \sum \frac{a^{2}}{ab+2ac+3ad}\geq \frac{\left ( a+b+c+d \right )^{2}}{4\left ( ab+bc+cd+da+ac+bd \right )}$( BĐT CAUCHY-SCHWARZ)
mà : $\left ( a+b+c+d \right )^{2}= 2(ab+bc+cd+da+ac+bd)+a^{2}+b^{2}+c^{2}+d^{2}\geq \frac{8}{3}\left ( ab+bc+cd+da+ac+bd \right )$
suy ra VT$\geq \frac{2}{3}$ (đpcm)

[lovemoon]

$\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\geq 1$
gọi vế trái là M
Ta có :$M=\sum \frac{a^{2}}{ab+2ac}$
$3\left ( ab+bc+ca \right )M\geq \left ( a+b+c \right )^{2}$ (BĐT cauchy schwarz)
mà$3\left ( ab+bc+ca \right )\leq \left ( a+b+c \right )^{2}$
suy ra $M\geq 1$ (đpcm)

[banhgaongonngon]

• $\frac{a^{8}+b^{8}+c^{8}}{a^{3}b^{3}c^{3}}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ (a,b,c>0)

Áp dụng nhiều lần bất đẳng thức $x^{2}+y^{2}+z\geq xy+yz+zx,\forall x,y,z\in \mathbb{R}$ ta được

$\sum a^{8}\geq \sum a^{4}b^{4}\geq a^{2}b^{4}c^{2}\geq \sum a^{2}b^{3}c^{3}=\prod (a^{3}).\sum \frac{1}{a}$

Do $a,b,c>0$ nên

$\frac{\sum a^{8}}{\prod a^{3}}\geq \sum \frac{1}{a}$