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Ta có : \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\ge9\)
\(\Leftrightarrow\frac{a+1}{a}.\frac{b+1}{b}\ge9\Leftrightarrow ab+a+b+1\ge9ab\) ( vì \(ab>0\) )
\(\Leftrightarrow a+b+1\ge8ab\Leftrightarrow2\ge8ab\) ( vì \(a+b=1\) )
\(\Leftrightarrow1\ge4ab\Leftrightarrow\left(a+b\right)^2\ge4ab\) ( Vì \(a+b=1\) ) \(\Leftrightarrow\left(a-b\right)^2\ge0\left(2\right)\)
BĐT ( 2 ) đúng , mà các phép biến đổi trê tương đương , vây BĐT ( 1 ) được chứng minh . Xảy ra đẳng thức khi và chỉ khi \(a=b\)
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)
\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có ít nhất 2 số cùng phía so với 1
Không mất tính tổng quát, giả sử đó là a và b
\(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\)
\(\Leftrightarrow ab+1\ge a+b\)
\(\Leftrightarrow2\left(ab+1\right)\ge\left(a+1\right)\left(b+1\right)\)
\(\Rightarrow\dfrac{2}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{2}{2\left(ab+1\right)\left(c+1\right)}=\dfrac{1}{\left(ab+1\right)\left(c+1\right)}=\dfrac{1}{\left(\dfrac{1}{c}+1\right)\left(c+1\right)}=\dfrac{c}{\left(c+1\right)^2}\)
Lại có:
\(\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{a}{b}}+1.1\right)^2}+\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{b}{a}}+1\right)^2}\ge\dfrac{1}{\left(ab+1\right)\left(\dfrac{a}{b}+1\right)}+\dfrac{1}{\left(ab+1\right)\left(\dfrac{b}{a}+1\right)}=\dfrac{1}{ab+1}\)
\(\Rightarrow P\ge\dfrac{1}{ab+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}=\dfrac{1}{\dfrac{1}{c}+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}\)
\(\Rightarrow P\ge\dfrac{c}{c+1}+\dfrac{c+1}{\left(c+1\right)^2}=\dfrac{c\left(c+1\right)+c+1}{\left(c+1\right)^2}=\dfrac{\left(c+1\right)^2}{\left(c+1\right)^2}=1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
Bài này đã có ở đây:
Cho abc=1CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\) - Hoc24
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 )
a) theo định lý côsi :
\(\dfrac{a}{b}\)+\(\dfrac{b}{a}\)luôn >=2 với mọi a, b , a.b > 0
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