Question

chứng minh bất đẳng thức

A=(a+b)(\(\dfrac{1}{a}\)+\(\dfrac{1}{b}\))\(\ge\) 4
help me!!TT

b) B=\(\dfrac{a+b}{c}\)+\(\dfrac{b+c}{a}\)+\(\dfrac{c+a}{b}\)\(\ge\)6 ( a,b,c > 0)

Answer 1

a)ĐK: a>0 b>0 nhé bạn đề thiếu

(a-b)²\(\ge\)0

<=>a²+b²\(\ge\)2ab

<=>a²+2ab+b²\(\ge\)4ab

<=>(a+b)²\(\ge\)4ab

<=>\(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)

<=>\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

Dấu "=" xảy ra <=> (a-b)²=0<=>a=b

=>A\(\ge\)\(\left(a+b\right)\dfrac{4}{a+b}=4\)(đpcm)

b)\(B=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{a+c}{b}=\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)

Áp dụng bất đẳng thức cosi x+y\(\ge\)2\(\sqrt{xy}\)cho 2 số dương x;y ta có:

\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{ac}{ca}}=2\)

\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\)

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\)

Dấu "=" xảy ra khi và chỉ khi:\(\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{c}{a}\\\dfrac{b}{c}=\dfrac{c}{b}\\\dfrac{a}{b}=\dfrac{b}{a}\end{matrix}\right.\)\(\Leftrightarrow\)a=b=c

=>B\(\ge2+2+2=6\)(đpcm)

Answer 2

Xuân Tuấn Trịnh

nếu không cần a>0; b>0

thì sao bạn