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Cách khác:
Đặt \(A=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\)
\(A=\left(1+\dfrac{a+b}{a}\right)\left(1+\dfrac{a+b}{b}\right)\)
\(A=\left(2+\dfrac{b}{a}\right)\left(2+\dfrac{a}{b}\right)\)
\(A=4+2\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+1\)
\(A\ge4+2\cdot2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}+1=9\left(AM-GM\right)\left(đpcm\right)\)
( 1 + \(\dfrac{1}{a}\))\(\left(1+\dfrac{1}{b}\right)\) ≥ 9
Biến đổi VT Ta có : VT = \(\dfrac{a+1}{a}.\dfrac{b+1}{b}\)
= \(\dfrac{2a+b}{a}.\dfrac{2b+a}{b}\)
=\(\left(2+\dfrac{b}{a}\right)\left(2+\dfrac{a}{b}\right)\)
= 4 + \(\dfrac{2a}{b}+\dfrac{2b}{a}+\dfrac{b}{a}.\dfrac{a}{b}\)
= 5 + 2( \(\dfrac{a}{b}+\dfrac{b}{a}\) ) ( *)
Áp dụng BĐT : \(\dfrac{x}{y}+\dfrac{y}{x}\) ≥ 2( x > 0 ; y > 0) ( ** )
Từ ( * ; **) ⇒ 5 + 2( \(\dfrac{a}{b}+\dfrac{b}{a}\) ) ≥ 5 + 4 = 9 ( đpcm )
Đặt A = \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\)
A = \(\left(1+\frac{a+b}{a}\right)\left(1+\frac{a+b}{b}\right)\)(Vì a + b = 1)
A = \(\left(2+\frac{b}{a}\right)\left(2+\frac{a}{b}\right)\)
A = \(4+\frac{2a}{b}+\frac{2b}{a}+1\)
A = \(5+2\left(\frac{a}{b}+\frac{b}{a}\right)\)
Vì a, b dương nên áp dụng BĐT Cô - si cho 2 số dương, ta được :
\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{ab}{ba}}\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}\ge2.1=2\)
\(\Leftrightarrow2\left(\frac{a}{b}+\frac{b}{a}\right)\ge4\)
\(\Leftrightarrow5+2\left(\frac{a}{b}+\frac{b}{a}\right)\ge4+5\)
\(\Leftrightarrow A\ge9\)
Dấu bằng xảy ra \(\Leftrightarrow\)a = b > 0
Vậy \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\ge9\)với a, b là các số dương và a + b = 1
Tớ quên. Dấu bằng xảy ra
\(\Leftrightarrow\hept{\begin{cases}a=b>0\\a+b=1\end{cases}}\)
\(\Leftrightarrow a=b=\frac{1}{2}\)
Ta có \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\ge9\) (1)
\(\Leftrightarrow\frac{a+1}{a}.\frac{b+1}{b}\ge9\)
\(\Leftrightarrow ab+a+b+1\ge9ab\) (vì ab > 0)
\(\Leftrightarrow a+b+1\ge8ab\Leftrightarrow2\ge8ab\) (vì a + b = 1)
\(\Leftrightarrow1\ge4ab\Leftrightarrow\left(a+b\right)^2\ge4ab\) (vì a + b = 1)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) (2)
Bất đẳng thức (2) đúng, mà các phép biến đổi trên tương đương, vậy bất đẳng thức (1) được chưng minh.
1+1/a= 1+ (a+b)/a = 2+b/a
tương tự: 1+1/b= 2+a/b
nhân 2 đa thức với nhau đc : 5+2a/b+2b/a=5+2(a/b+b/a)
áp dụng bđt cô si a/b+b/a >=2 =) 5+2(a/b+b/a)>=9 (dấu = xảy ra khi a-b=1/2)
a) theo định lý côsi :
\(\dfrac{a}{b}\)+\(\dfrac{b}{a}\)luôn >=2 với mọi a, b , a.b > 0
BDT
\(x+\dfrac{1}{x}=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)^2+2\ge2\)
nhân PP vào là ra
\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3+2+2+2=9\)
Theo BĐT Cauchy:
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\)
Đẳng thức xảy ra \(\Leftrightarrow a=b=c\)
thử bài bất :D
Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)
Hoàn toàn tương tự:
\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)
\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)
Cộng (*),(**),(***) vế theo vế ta được:
\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)
Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )
Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=1
1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D
Giải:
Áp dụng BĐT Cô si cho 3 số dương ta được:
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)
Nhân theo vế 2 BĐT trên ta có:
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\) (Đpcm)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Nếu đề là \(a,b,c\ge0\) thì làm như sau:
Áp dụng bất đẳng thức Cauchy ta có:
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(a+b+c\right).\dfrac{9}{\left(a+b+c\right)}=9\)
Đẳng thức xảy ra khi a = b = c
\(\Rightarrowđpcm\)
Đặt \(P=\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(abc\right)^2}{a^3\left(b+c\right)}+\dfrac{\left(abc\right)^2}{b^3\left(c+a\right)}+\dfrac{\left(abc\right)^2}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ca\right)^2}{b\left(c+a\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\)
\(P\ge\dfrac{\left(bc+ca+ab\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\) (BĐT B.C.S)
\(=\dfrac{ab+bc+ca}{2}\) \(\ge\dfrac{3\sqrt[3]{abbcca}}{2}=\dfrac{3}{2}\) (do \(abc=1\)).
ĐTXR \(\Leftrightarrow a=b=c=1\)
a)\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
=\(\dfrac{a}{a}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{c}+\dfrac{c}{a}+\dfrac{c}{b}\)
=\(1+1+1+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\)
=3+\(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\)
áp dụng BĐT cô si ta có
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}\)
⇔ \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
cmtt ta có \(\dfrac{b}{c}+\dfrac{c}{b}\ge2\); \(\dfrac{a}{c}+\dfrac{c}{a}\ge2\)
=> 3+\(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge9\)
=> \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\left(đpcm\right)\)
a)Áp dụng bđt AM-GM cho 3 số không âm ta có:
\(a+b+c\ge3\sqrt[3]{abc}\)
TT\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)
Nhân vế theo vế ta có:\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}\cdot3\sqrt[3]{\dfrac{1}{abc}}=9\left(đpcm\right)\)
b)\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\)
Svac-xo:
\(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
Lại có:\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)(tự cm)
\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)