Câu 2)

Ta có \(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{3}\)

\(\Rightarrow\frac{b+1+a+1}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)

Ta có \(a+b=1\)

\(\Rightarrow\frac{3}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)

\(\Rightarrow\frac{3}{\left(a+1\right)b+a+1}\ge\frac{4}{3}\)

\(\Rightarrow\frac{3}{ab+b+a+1}\ge\frac{4}{3}\)

Ta có \(a+b=1\)

\(\Rightarrow\frac{3}{ab+2}\ge\frac{4}{3}\)

\(\Leftrightarrow9\ge4\left(ab+2\right)\)

\(\Rightarrow9\ge4ab+8\)

\(\Rightarrow1\ge4ab\)

Do \(a+b=1\Rightarrow\left(a+b\right)^2=1\)

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

\(\Rightarrow a^2+2ab+b^2\ge4ab\)

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

\(\Rightarrow\left(a-b\right)^2\ge0\) (đpcm )

Câu 3)

Ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)

Mà \(a+b+c=1\)

\(\Rightarrow\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\ge9\)

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

Áp dụng bất đẳng thức Cô-si

\(\Rightarrow\left\{\begin{matrix}a+b+c\ge3\sqrt[3]{abc}\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\end{matrix}\right.\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc}\sqrt[3]{\frac{1}{abc}}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9.\sqrt[3]{\frac{abc}{abc}}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (điều này luôn luôn đúng)

\(\Rightarrow\) ĐPCM

Câu 1: a)

b) Áp dụng Bđt Holder ta có:

\(\Rightarrow9\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)^3\)

\(\Rightarrow\frac{a^3+b^3+c^3}{3}\ge\frac{\left(a+b+c\right)^3}{27}=\left(\frac{a+b+c}{3}\right)^3\)(đpcm)

Dấu = khi a=b=c

Câu 2:

Áp dụng Bđt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)ta có:

\(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{a+b+1+1}=\frac{4}{3}\)(Đpcm)

Dấu = khi \(a=b=\frac{1}{2}\)

Câu 3:

Áp dụng Bđt \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)ta có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=9\left(a+b+c=1\right)\)(Đpcm)

Dấu = khi \(a=b=c=\frac{1}{3}\)

Câu 4: nghĩ sau

\(\dfrac{1}{\left(a+b\right)^3}\left(\dfrac{1}{a^3}+\dfrac{1}{a^3}\right)+\dfrac{3}{\left(a+b\right)^4}+\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)+\dfrac{6}{\left(a+b\right)^5}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

\(=\dfrac{1}{\left(a+b\right)^3}\cdot\dfrac{b^3+a^3}{a^3b^3}+\dfrac{3}{\left(a+b\right)^4}\cdot\dfrac{b^2+a^2}{a^2b^2}+\dfrac{6}{\left(a+b\right)^5}\cdot\dfrac{b+a}{ab}\)

\(=\dfrac{1}{\left(a+b\right)^3}\cdot\dfrac{\left(b+a\right)\left(a^2-ab+a^2\right)}{a^3b^3}+\dfrac{3\left(b^2+a^2\right)}{a^2b^2\cdot\left(a+b\right)^4}\cdot\dfrac{6}{\left(a+b\right)^4}\cdot\dfrac{1}{ab}\)

\(=\dfrac{1}{\left(a+b\right)^2}\cdot\dfrac{b^2-ab+a^2}{a^3b^3}+\dfrac{3b^2+3a^2}{a^2b^2\cdot\left(a+b\right)^4}+\dfrac{6}{ab\left(a+b\right)^4}\)

\(=\dfrac{b^2-ab+a^2}{a^3b^3\cdot\left(a+b\right)^2}+\dfrac{3b^2+3a^2}{a^2b^2\cdot\left(a+b\right)^4}+\dfrac{6}{ab\cdot\left(a+b\right)^4}\)

\(=\dfrac{\left(a+b\right)^2\cdot\left(b^2-ab+a^2\right)+ab\left(3b^2+3a^2\right)+6a^2b^2}{a^3b^3\cdot\left(a+b\right)^4}\)

\(=\dfrac{\left(a^2+2ab+b^2\right)\left(b^2-ab+a^2\right)+3ab^3+3a^3b+6a^2b^2}{a^3b^3\cdot\left(a+b\right)^4}\)

\(=\dfrac{a^2b^2-a^3b+a^4+2ab^3-2a^2b^2+2a^3b+b^4-ab^3+a^2b^2+3ab^3+3a^2b+6a^2b^2}{a^3b^3\cdot\left(a+b\right)^4}\)

\(=\dfrac{6a^2b^2+4a^3b+a^4+4ab^3+b^4}{a^3b^3\cdot\left(a+b\right)^4}\)

Ta chứng minh BĐT sau với các số dương:

\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Thật vậy, BĐT tương đương: \(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)

Áp dụng:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ; \(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) ; \(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)

Cộng vế với vế:

\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)

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

b.

Ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)

\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\Rightarrow\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)

\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)

Cộng vế với vế (1); (2) và (3):

\(\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Làm tạm một câu rồi đi chơi, lát làm cho.

4)

Áp dụng bất đẳng thức Cauchy-Schwarz :

\(VT\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1}=9\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)

2/ Cô: \(\frac{2a}{b}+\frac{b}{c}\ge3\sqrt[3]{\frac{a.a.b}{b.b.c}}=3\sqrt[3]{\frac{a^3}{abc}}=\frac{3a}{\sqrt[3]{abc}}\)

Tương tự hai BĐT còn lại và cộng theo vế thu được:

\(3.VT\ge3.VP\Rightarrow VT\ge VP^{\left(Đpcm\right)}\)

Đẳng thức xảy ra khi a = b= c

Ta co:a-b=15

=>2(a-b)=30 hay 2a-2b=30

Co:\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\)

\(hay\frac{2a}{4}=\frac{2b}{3}=\frac{3c}{4}\)va 2a-2b=30

Ap dung tinh chat cua day ti so bang nhau ta co:

\(\frac{2a}{4}=\frac{2b}{3}=\frac{3c}{4}=\frac{2a-2b}{4-3}=\frac{30}{1}=30\)

Con lai la tu ban nhe

ko hieu hoi mik

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