kfcchicken98

bài 4

bđt tương đương $\frac{a^{2}}{b}-2a+b+\frac{b^{2}}{c}-2b+c+\frac{c^{2}}{a}-2c+a\geq \frac{4(a-b)^{2}}{a+b+c}$

tương đương $\frac{(a-b)^{2}}{b}+\frac{(b-c)^{2}}{c}+\frac{(c-a)^{2}}{a}\geq \frac{4(a-b)^{2}}{a+b+c}$

có $\frac{(a-b)^{2}}{b}+\frac{(b-c)^{2}}{c}+\frac{(c-a)^{2}}{a}\geq \frac{(|a-b|+|b-c|+|c-a|)^{2}}{a+b+c}\geq \frac{(2|a-b|)^{2}}{a+b+c}=\frac{4(|a-b|^{2})}{a+b+c}$

cũng có thể giải theo cách này

bđt tương đương $\sum \frac{(a+b)(c+a)}{b+c}\geq 2$

đặt (a+b)=x; (b+c)=y; (c+a)=z

cần CM $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\geq 2$

đặt $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}=P$

$(xyz+xyz+xyz)=S$

có $P.S\geq (xy+yz+xz)^{2}$

suy ra $P\geq \frac{(xy+yz+xz)^{2}}{3xyz}\geq \frac{3xyz(x+y+z)}{3xyz}=x+y+z=2$

Lời giải. Ta có $a+b+c= \frac 1a + \frac 1b + \frac 1c+8 \ge \frac{9}{a+b+c}+8$.

Vì $a,b,c$ dương nên $a+b+c$ dương. Do đó $$(a+b+c)^2 \ge 9+8(a+b+c) \Leftrightarrow (a+b+c-4)^2 \ge 25 \Leftrightarrow a+b+c \ge 9$$

Ta có $ab+bc+ca \le \frac{(a+b+c)^2}{3}= \frac{81}{3}=27$.

Dấu đẳng thức xảy ra khi và chỉ khi $a=b=c=3$. $\blacksquare$

ngược dấu 

1. cho x,y,z dương, CM

$x^{3}(y^{2}+z^{2})^{2}+y^{3}(z^{2}+x^{2})^{2}+z^{3}(x^{2}+y^{2})^{2}\geq xyz(xy(x+y)^{2}+yz(y+z)^{2}+xz(x+z)^{2})$ (USA TST 2009)

2. cho a,b,c dương, abc=1

Tìm Min $\frac{1}{a^{5}(b+2c)^{2}}+\frac{1}{b^{5}(c+2a)^{2}}+\frac{1}{c^{5}(a+2b)^{2}}$ (USA TST 2010)

$\lim_{x\rightarrow 0}\frac{\ln (1+3x)}{x}=\lim_{x\rightarrow 0}3\frac{\ln (1+3x)}{3x}=3$

có $\frac{a^{3}}{b^{3}}+1+1\geq \frac{3a}{b}$

$\frac{a^{3}}{c^{3}}+1+1\geq \frac{3a}{c}$

$\frac{b^{3}}{a^{3}}+1+1\geq \frac{3b}{a}$

$\frac{b^{3}}{c^{3}}+1+1\geq \frac{3b}{c}$

$\frac{c^{3}}{a^{3}}+1+1\geq \frac{3c}{a}$

$\frac{c^{3}}{b^{3}}+1+1\geq \frac{3c}{b}$

suy ra $(a^{3}+b^{3}+c^{3})(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}})+9\geq 3(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b})$

do $\frac{3}{2}(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}) =\frac{3}{2}(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{b})\geq \frac{3}{2}(3+3)=9$

suy ra $(a^{3}+b^{3}+c^{3})(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}})\geq \frac{3}{2}(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b})$

cách 1; đặt $S=x(1-x)+y(1-y)$

suy ra $P^{2}S\geq (x+y)^{3}$(Holder)

suy ra $P^{2}\geq \frac{(x+y)^{3}}{x(1-x)+y(1-y)}=\frac{1}{2xy}\geq 2$

suy ra P min=$\sqrt{2}$

cach 2

$\frac{x}{\sqrt{1-x}}+\frac{y}{\sqrt{1-y}}=\frac{x^{2}}{\sqrt{x}\sqrt{xy}}+\frac{y^{2}}{\sqrt{y}\sqrt{xy}}\geq \frac{(x+y)^{2}}{\sqrt{x}\sqrt{xy}+\sqrt{y}\sqrt{xy}}\geq \frac{1}{\sqrt{(x+y)2xy}}= \frac{1}{\sqrt{2xy}}\geq \sqrt{2}$

giải:$\frac{a}{\sqrt{bc(1+a^{2})}}=\frac{a\sqrt{bc}}{\sqrt{bc(bc+a^{2}bc})}=\frac{a}{\sqrt{bc+a^{2}+ab+ac}}=\frac{a}{\sqrt{(a+b)(a+c)}}\leq \frac{a}{2(a+b)}+\frac{a}{2(a+c)}$

tương tự,suy ra VT$\leq \frac{1}{2}(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{b+a}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{c}{b+c})=\frac{3}{2}$

dấu $=$ khi a=b=c=$\sqrt{3}$

câu b
bđt tương đương $\frac{a^{2}c}{a+b}+\frac{b^{2}a}{b+c}+\frac{c^{2}b}{a+c}\geq 1/2(ab+bc+ca)$

$\frac{a^{2}c}{a+b}+\frac{b^{2}a}{b+c}=\frac{a^{2}c^{2}}{ac+bc}+\frac{b^{2}a^{2}}{ab+ac}+\frac{c^{2}b^{2}}{bc+ab}\geq \frac{(ab+bc+ca)^{2}}{2(ab+bc+ca)}=\frac{ab+bc+ca}{2}$
P/S phải là lớn hơnhơn hoặc bằng $\frac{1}{2}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

câu B

1. áp dụng bđt Cauchy Schwarz: $(a+b+c)(\frac{a}{(b+c)^{2}}+\frac{b}{(c+a)^{2}}+\frac{c}{(a+b)^{2}})\geq (\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})^{2}$ 

do$(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})^{2}\geq \frac{9}{4}$ bđt nesbitt

suy ra VT$\geq \frac{9}{4(a+b+c)}$

bài 1

$P^{2}\leq 3(\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{1}{1+z^{2}})$

giờ cần chứng minh$3(\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{1}{1+z^{2}})\leq \frac{9}{4}$

bđt tương đương $\frac{x^{2}}{1+x^{2}}+\frac{y^{2}}{1+y^{2}}+\frac{z^{2}}{1+z^{2}}\geq \frac{9}{4}$

tương đương $\frac{1}{1+\frac{1}{x^{2}}}+\frac{1}{1+\frac{1}{y^{2}}}+\frac{1}{1+\frac{1}{z^{2}}}\geq \frac{9}{4}$

có $\frac{1}{1+\frac{1}{x^{2}}}+\frac{1}{1+\frac{1}{y^{2}}}+\frac{1}{1+\frac{1}{z^{2}}}\geq \frac{9}{3+\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}}$

do x+y+z=xyz, suy ra $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1$

suy ra $\frac{9}{3+\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}}=\frac{9}{\frac{4}{3}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}}=\frac{27(x+y+z)^{2}}{4(xy+yz+xz)^{2}}\geq \frac{81}{4(xy+yz+xz)}$

$1=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\geq \frac{9}{xy+yz+xz}$

suy ra $\frac{81}{4(xy+yz+xz)}\geq \frac{9}{4}$ đpcm

bài 2, trâu bò tí

P= $\frac{x^{2}}{\sqrt{x}\sqrt{x^{2}y+xz}}+\frac{y^{2}}{\sqrt{y}\sqrt{xy^{2}+yz}}+\frac{y^{2}}{\sqrt{y}\sqrt{y^{2}z}+xy}+\frac{z^{2}}{\sqrt{z}\sqrt{yz^{2}+xz}}+\frac{x^{2}}{\sqrt{x}\sqrt{x^{2}z+yx}}+\frac{z^{2}}{\sqrt{z}\sqrt{xz^{2}+yz}}\geq \frac{4(x+y+z)^{2}}{\sum \sqrt{x}\sqrt{x^{2}+xy}}\geq \frac{4}{\sqrt{2((\frac{2}{9}(a+b+c)^{3}+\frac{2(a+b+c)^{2}}{3}}}=\frac{4}{\sqrt{2(\frac{2}{9}+\frac{2}{3})}}=3$

tạm thời chưa có cách ngắn hơn

giải: có 2 cách làm bài này, cách 1 là thay $b^{2}+c^{2}=1-a^{2}$ và tương tự rồi chứng minh

cách 2: $\frac{a}{b^{2}+c^{2}}+\frac{b}{c^{2}+a^{2}}+\frac{c}{a^{2}+b^{2}}=\frac{a^{2}}{ab^{2}+c^{2}a}+\frac{b^{2}}{bc^{2}+a^{2}b}+\frac{c^{2}}{a^{2}c+b^{2}c}\geq \frac{(a+b+c)^{2}}{ab(a+b)+bc(c+b)+ca(c+a)}$

do $2(a+b+c)^{3}\geq 9(ab(a+b)+bc(c+b)+ca(a+c))$

suy ra $P\geq \frac{(a+b+c)^{2}}{\frac{2}{9}(a+b+c)^{3}}=\frac{9}{2(a+b+c)}$

mặt khác $1=a^{2}+b^{2}+c^{2}\geq \frac{(a+b+c)^{2}}{3}$

suy ra $a+b+c\leq \sqrt{3}$

suy ra $P\geq \frac{9}{2\sqrt{3}}=\frac{3\sqrt{3}}{2}$

cho a,b,c là các số dương

cm: $(a+b)^{2}+(a+b+4c)^{2}\geq \frac{100abc}{a+b+c}$

cho x,y,z >0; xyz=1

CM $\frac{1}{\sqrt{1+3x}}+\frac{1}{\sqrt{1+3y}}+\frac{1}{\sqrt{1+3z}}\geq \frac{3}{2}$