kfcchicken98

bài 4

đặt $x=\frac{b}{a}; y=\frac{c}{b};z=\frac{a}{c}$, suy ra xyz=1

ta có $\frac{1}{(x+1)^{3}}+\frac{1}{(x+1)^{3}}+\frac{1}{8}\geq \frac{3}{2}\frac{1}{(x+1)^{2}}$

tương tự $\sum \frac{1}{(1+x)^{3}}\geq \frac{3}{2}(\sum \frac{1}{(1+x)^{2}})-\frac{3}{8}$

giờ cần CM $\sum \frac{1}{(1+x)^{2}}\geq \frac{3}{4}$

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

có $\frac{1}{xy+1}+\frac{1}{(1+z)^{2}}=\frac{z}{z+1}+\frac{1}{(1+z)^{2}}=\frac{z^{2}+z+1}{(z+1)^{2}}$

giả sử z là max (x,y,z), suy ra $z\geq 1$

xét đạo hàm $\frac{z^{2}+z+1}{z^{2}+2z+1}$ >0

suy ra $f(z)\geq f(1)=\frac{3}{4}$ đpcm

bài 3

do bđt thuần nhất nên chuẩn hóa a+b+c=1

$P\geq \frac{(a+b+c)^{3}}{abc}+\frac{(ab+bc+ca)^{3}}{{\frac{1}{27}}(a+b+c)^{6}}=\frac{(a+b+c)^{3}}{abc}+\frac{27(ab+bc+ca)^{3}}{(a+b+c)^{6}}$

ta có $\frac{(a+b+c)^{3}}{27abc}+\frac{(a+b+c)^{3}}{27abc}+\frac{27(ab+bc+ca)^{3}}{(a+b+c)^{6}}\geq \frac{(ab+bc+ca)}{\sqrt[3]{(abc)^{2}}}\geq \frac{9(ab+bc+ca)}{(a+b+c)^{2}}$

có $\frac{9(ab+bc+ca)}{(a+b+c)^{2}}+\frac{3(a+b+c)^{3}}{27abc}\geq 2\sqrt{\frac{(a+b+c)(ab+bc+ca)}{abc}}\geq 6$

có $\frac{22(a+b+c)^{3}}{27abc}\geq 22$

suy ra P$\geq 22+6=28$

đặt P=$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}$

S= $a(b+c)+b(c+a)+c(a+b)$

có $P^{2}S\geq (a+b+c)^{3}$

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

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

bài 2

$\sum \frac{2n-1}{2^{n}}=\sum \frac{2n}{2^{n}}-\sum \frac{1}{2^{n}}$

có$\sum \frac{1}{2^{n}}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$

$\sum \frac{2n}{2^{n}}=2\sum \frac{n}{2^{n}}=2\sum \frac{2n-(n+1)+1}{2^{n}}=2\sum (\frac{n}{2^{n-1}}-\frac{n+1}{2^{n}}+\frac{1}{2^{n}})=2(1-\frac{n+1}{2^{n}})+2$

suy ra $\lim u_{n}=2(1-0+1)-1=3$

ta có $M=\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}$

$\frac{1}{1-ab}-1=\frac{ab}{1-ab}$

suy ra M-3=$\frac{ab}{1-ab}+\frac{bc}{1-bc}+\frac{ca}{1-ac}$

ta có $2M-6=\frac{4ab}{2-2ab}+\frac{4bc}{2-2bc}+\frac{4ca}{2-2ca}=\frac{4ab}{(a-b)^{2}+c^{2}+1}+\frac{4bc}{(b-c)^{2}+1+a^{2}}+\frac{4ca}{(c-a)^{2}+1+b^{2}}\leq \frac{(a+b)^{2}}{a^{2}+c^{2}+b^{2}+c^{2}}+\frac{(b+c)^{2}}{a^{2}+b^{2}+a^{2}+c^{2}}+\frac{(c+a)^{2}}{c^{2}+b^{2}+a^{2}+b^{2}}\leq \frac{a^{2}}{a^{2}+c^{2}}+\frac{b^{2}}{b^{2}+c^{2}}+\frac{b^{2}}{a^{2}+b^{2}}+\frac{c^{2}}{a^{2}+c^{2}}+\frac{c^{2}}{b^{2}+c^{2}}+\frac{a^{2}}{a^{2}+b^{2}}=3$

suy ra $2M-6\leq 3$

suy ra $M\leq \frac{9}{2}$

$P\geq \frac{9}{(a+b+c)^{2}}+\frac{2011}{3}=\frac{2014}{3}$

bài 1

ta có$\sqrt{\frac{a}{b+c+2a}}+\sqrt{\frac{b}{a+c+2b}}+\sqrt{\frac{c}{a+b+2c}}\leq \sqrt{3(\frac{a}{b+c+2a}+\frac{b}{a+c+2b}+\frac{c}{a+b+2c})}\leq \sqrt{\frac{3}{4}(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{c}{b+c})}=\sqrt{\frac{9}{4}}=\frac{3}{2}$

bài 2

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

có $\frac{a^{2}-ab+b^{2}}{ab}+\frac{b^{2}+c^{2}-bc}{bc}+\frac{a^{2}-ac+c^{2}}{ac}\geq \frac{a^{2}+b^{2}}{2ab}+\frac{b^{2}+c^{2}}{2bc}+\frac{c^{2}+a^{2}}{2ac}=\frac{a}{2b}+\frac{b}{2a}+\frac{b}{2c}+\frac{c}{2b}+\frac{c}{2a}+\frac{a}{2c}$

ta có $2(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})\leq \frac{1}{2}(\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b})$ đpcm

câu 1

bđt tương đương $\frac{1}{\sqrt[4]{a}}\sqrt[4]{\sqrt{3}+6\sqrt{3}ab}+\frac{1}{\sqrt[4]{b}}\sqrt[4]{\sqrt{3}+6\sqrt{3}bc}+\frac{1}{\sqrt[4]{c}}\sqrt[4]{\sqrt{3}+6\sqrt{3}ac}\leq \frac{1}{abc}$

có $\sum \frac{1}{\sqrt[4]{a}}\sqrt[4]{(\sqrt{3}+6\sqrt{3}ab)}\leq (\sqrt{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}})\sqrt{\sqrt{\sqrt{3}+6\sqrt{3}ab}+\sqrt{\sqrt{3}+6\sqrt{3}bc}+\sqrt{\sqrt{3}+6\sqrt{3}ca}}\leq \sqrt[4]{\frac{3}{abc}}\sqrt[4]{27\sqrt{3}}$

giờ cần CM $\sqrt[4]{\frac{3}{abc}}\sqrt[4]{27\sqrt{3}}\leq \frac{1}{abc}$

tương đương ${\frac{3}{abc}}{27\sqrt{3}}\leq \frac{1}{(abc)^{4}}$

tương đương $\frac{1}{81\sqrt{3}}\geq (abc)^{3}$

ta có $abc\leq (\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{3})^{3}\leq \frac{(\sqrt{3(ab+bc+ca)})^{3}}{27}=\frac{(\sqrt{3})^{3}}{27}=\frac{\sqrt{3}}{9}=\frac{1}{3\sqrt{3}}$

suy ra $(abc)^{3}\leq (\frac{1}{3\sqrt{3}})^{3}=\frac{1}{27.3\sqrt{3}}=\frac{1}{81\sqrt{3}}$ đpcm 

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

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

áp dụng bđt minkowski $\sqrt{1+\frac{1}{a^{2}}}+\sqrt{1+\frac{1}{b^{2}}}+\sqrt{1+\frac{1}{c^{2}}}\geq \sqrt{9+(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}\geq \sqrt{9+\frac{3}{ab}+\frac{3}{bc}+\frac{3}{ca}}$

do a+b+c=abc, suy ra $\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1$

suy ra $\sqrt{9+\frac{3}{ab}+\frac{3}{bc}+\frac{3}{ca}}=\sqrt{9+3}=2\sqrt{3}$

chuẩn hóa $a+b+c=1$

bđt tuong đương $\frac{a(1-a)}{(1-a)^{2}+a^{2}}+\frac{b(1-b)}{(1-b)^{2}+b^{2}}+\frac{c(1-c)}{(1-c)^{2}+c^{2}}$

có $\frac{a(1-a)}{(1-a)^{2}+a^{2}}=\frac{a(1-a)}{2a^{2}-2a+1}=\frac{a(1-a)}{1-2a(1-a)}\leq \frac{a(1-a)}{1-\frac{(a+1)^{2}}{4}}=\frac{a(1-a)}{(1-\frac{a+1}{2})(1+\frac{a+1}{2})}=\frac{4a(1-a)}{(1-a)(a+3)}=\frac{4a}{a+3}=4-\frac{12}{a+3}$

suy ra $H\leq 12-12(\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3})\leq 12-12\frac{9}{10}=\frac{6}{5}$

suy ra max= $\frac{6}{5}$

giải, bđt tương đương $2(1+a^{2})(1+b^{2})\leq (1+ab)(1+a^{2}+1+b^{2})$

$2(1+a^{2}+b^{2}+a^{2}b^{2})\leq (1+ab)(1+a^{2}+1+b^{2})$

tương đương $2(1+a^{2}+b^{2}+a^{2}b^{2})\leq 1+a^{2}+1+b^{2}+ab+a^{3}b+ab+ab^{3}$

tương đương $a^{2}+b^{2}+2a^{2}b^{2}\leq 2ab+ab(a^{2}+b^{2})$

tương đương $2ab(1-ab)+(a^{2}+b^{2})(ab-1)\geq 0$

tương đương $(ab-1)(a^{2}-2ab+b^{2})\geq 0$

do a,b>1, suy ra ab-1>0; $(a-b)^{2}\geq 0$ , suy ra đpcm

không biết bạn có cần nữa không, mình xin đưa ra 1 cách giúp bạn

có $x_{n+1}-1=\frac{x_{n}-1}{x_{n}+2}$

$x_{n+1}+1=\frac{3(x_{n}+1)}{x_{n}+2}$

suy ra $\frac{x_{n+1}+1}{x_{n+1}-1}= \frac{3(x_{n}+1)}{x_{n}-1}$

có $\frac{(x_{n}+1)}{x_{n}-1}=\frac{3(x_{n-1}+1)}{x_{n-1}-1}=...=\frac{3^{n}(x_{0}+1)}{x_{0}-1}=3^{n+1}$

từ đó đưa ra được $x_{n}=\frac{3^{n+1}+1}{3^{n+1}-1}$

có $x_{n}=\frac{3^{n+1}+1}{3^{n+1}-1}=1+\frac{2}{3^{n+1}-1}$

suy ra $x_{1}+x_{2}+...+x_{2008}=2008+\sum_{n=1}^{2008}\frac{2}{3^{^{n+1}}-1}$

giờ cần CM $\sum_{n=1}^{2008}\frac{2}{3^{^{n+1}}-1}< 1$

ta có $\sum \frac{2}{3^{n+1}-1}< \sum \frac{2}{\frac{1}{2}3^{n+1}}=\sum \frac{4}{3^{n+1}}$=$\4\frac{\frac{1}{9}}{1-\frac{1}{3}}=\frac{4}{6}< 1$

có $\sum_{n=1}^{2008}\frac{2}{3^{n+1}-1}< \sum \frac{2}{{3^{n+1}}-1}< 1$ đpcm

khi a=2; n=3

bđt tương đương $x^{3}y^{3}(x^{3}+y^{3})\leq 2$

tương đương $x^{3}y^{3}(x+y)(x^{2}-xy+y^{2})\leq 2$

có $x^{3}y^{3}(x^{2}-xy+y^{2})\leq (\frac{(xy+xy+xy+x^{2}-xy+y^{2})}{4})^{4}=(\frac{(x+y)^{2}}{4})^{4}=1$

và x+y=2; suy ra đpcm

bài 4

$\sqrt{a^{2}+abc}=\sqrt{a}\sqrt{a+bc}=\sqrt{a}\sqrt{a^{2}+ab+ac+bc}=\sqrt{a}\sqrt{(a+b)(a+c)}$

suy ra $\sum \sqrt{a^{2}+abc}=\sum \sqrt{a}\sqrt{a+b}\sqrt{a+c}\leq \sqrt{\sum (a+b)(a+c)}\leq \sqrt{(\frac{2a+2b+2c}{3})^{2}}=\frac{2}{\sqrt{3}}$

$9\sqrt{abc}\leq 9\sqrt{(\frac{(a+b+c)}{3})^{3}}=\frac{3}{\sqrt{3}}$

suy ra Max= $\frac{3}{\sqrt{3}}+\frac{2}{\sqrt{3}}=\frac{5}{\sqrt{3}}$