a) \(\frac{\sqrt{2x^3}}{\sqrt{8x}}=\sqrt{\frac{2x^3}{8x}}=\frac{1}{2}x\)

b) \(\left(3-\sqrt{5}\right)\left(x+\sqrt{5}\right)=3^2-\left(\sqrt{5}\right)^2=9-5=4\)

c) \(\sqrt{\frac{3x^2y^4}{27}}=0\)

\(y\ne0\)

Thì \(\sqrt{\frac{3x^2y^4}{27}}=\frac{1}{3}xy^2\)

e) \(\frac{y}{x^2}\sqrt{\frac{36x^4}{y^2}}=\frac{y}{x^2}.\frac{6x^2}{\left|y\right|}=\frac{6y}{\left|y\right|}\)

Vì y < 0 nên \(\left|y\right|=-y\)

Vậy \(\frac{6y}{\left|y\right|}=\frac{6y}{-y}=-6\)

f) \(\frac{\sqrt{99999999}}{\sqrt{11111111}}=\sqrt{\frac{99999999}{11111111}}=\sqrt{9}=3\)

\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)    \(\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{a+b}+1\ge\frac{9}{2}\) 

\(\Leftrightarrow\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\ge\frac{9}{2}\) 

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge\frac{9}{2}\)

\(\Leftrightarrow2\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge9\) 

thật vậy\(2\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)\) =\(\left[\left(b+c\right)+\left(a+c\right)+\left(a+b\right)\right]\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge9\) (ÁP DỤNG BẤT ĐẲNG THỨC COSI) 

ĐẲNG THỨC CUỐI ĐÚNG SUY RA ĐẲNG THỨC ĐẦU ĐƯỢC CHỨNG MINH

Đặt A=\(\sqrt{x-1-2\sqrt{x-2}}=\sqrt{\left(\sqrt{x-2}\right)^2-2\sqrt{x-2}.1+1^2}\)=\(\sqrt{\left(\sqrt{x-2}-1\right)^2}\)

Do\(x\ge3\)nên A  =\(\sqrt{x-2}-1\)

CMR a+2b+c >= 4(1-a)(1-b)(1-c) - Bất đẳng thức và cực trị - Diễn đàn Toán học

bạn có thể giải giúp mình bài toán nay ko. giúp mình nha

\(\frac{1+\sqrt{3}}{\sqrt{3}-1}=\frac{\left(1+\sqrt{3}\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=2+\sqrt{3}\)

\(\frac{2}{\sqrt{2}-1}=\frac{2\sqrt{2}+2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=2\sqrt{2}+2=\sqrt{8}+2\)

\(\Rightarrow\frac{2}{\sqrt{2}-1}>\frac{1+\sqrt{3}}{\sqrt{3}-1}\)

\(\sqrt{12}-\sqrt{11}\)   bé hơn \(\sqrt{11}-\sqrt{10}\) 

Ta cần chứng minh bất đẳng thức phụ: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\frac{a+b}{ab}\ge\frac{4}{a+b}\)

\(\frac{\left(a+b\right)^2}{ab\left(a+b\right)}\ge\frac{4ab}{ab\left(a+b\right)} \)

\(\left(a+b\right)^2\ge4ab\)

\(a^2+2ab+b^2-4ab\ge0\)

\(\left(a-b\right)^2\ge0\)(luôn đúng)

Xét c+1 = a+b+c+c

Áp dụng bất đẳng thức trên, ta có:

\(\frac{ab}{c+1}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)

\(\frac{bc}{a+1}\le\frac{bc}{4}\left(\frac{1}{c+a}+\frac{1}{a+b}\right)\)

\(\frac{ca}{b+1}\le\frac{ca}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\)

Cộng vế theo vế, ta có: 

\(\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ca}{b+1}\le\frac{1}{4}\left(\frac{ab}{b+c}+\frac{ab}{c+a}+\frac{bc}{c+a}+\frac{bc}{a+b}+\frac{ca}{a+b}+\frac{ca}{b+c}\right)\)

\(\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ca}{b+1}\le\frac{1}{4}\left(\frac{ab+ca}{b+c}+\frac{ab+bc}{c+a}+\frac{bc+ca}{a+b}\right)\)

\(\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ca}{b+1}\le\frac{1}{4}\left(\frac{a\left(b+c\right)}{b+c}+\frac{b\left(a+c\right)}{c+a}+\frac{c\left(b+a\right)}{a+b}\right)\)

\(\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ca}{b+1}\le\frac{1}{4}\left(a+b+c\right)\)

\(\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ac}{b+1}\le\frac{1}{4}\)

=> Điều phải chứng minh

ta có với x,y>0 thì \(\left(x+y\right)^2\ge4xy\Rightarrow\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Rightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)(*) dấu "=" xảy ra khi x=y

áp dụng bđt (*) và do a+b+c=1 nên ta có

\(\frac{ab}{c+1}=\frac{ab}{\left(c+a\right)+\left(c+b\right)}\le\frac{ab}{4}\left(\frac{1}{c+a}+\frac{1}{c+b}\right)\)

tương tự ta có \(\frac{bc}{a+1}\le\frac{bc}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right);\frac{ca}{b+1}\le\frac{ca}{4}\left(\frac{1}{b+a}+\frac{1}{b+c}\right)\)

\(\Rightarrow\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ca}{b+1}\le\frac{1}{4}\left(\frac{ab+bc}{c+a}+\frac{ab+ca}{b+c}+\frac{bc+ca}{a+b}\right)=\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\)

\(\Rightarrow\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{c+a}{b+1}\le\frac{1}{4}\)

dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)