cho a,b,c là các số dương , chứng tỏ:
a)\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
b)\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
Cho ba số dương a,b,c chứng minh rằng:
(a + b + c)\(\left(\dfrac{1}{a}++\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
a,b,c là các số dương nên \(\left(a+b+c\right)>=3\cdot\sqrt[3]{abc}\)
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=3\cdot\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}\)
Do đó: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=3\cdot\sqrt[3]{abc}\cdot3\cdot\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\cdot\sqrt[3]{a\cdot b\cdot c\cdot\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\)
Với a,b,c là số dương chứng minh rằng :
a, \(\left(a+b\right)\times\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)
b, \(\left(a+b+c\right)\times\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
=> (a+b).\(\left(\dfrac{1}{b}+\dfrac{1}{b}\right)\ge\left(a+b\right).\dfrac{4}{a+b}=4\left(dpcm\right)\)
b)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+b+c}\)
=>\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(a+b+c\right).\dfrac{9}{a+b+c}=9\left(dpcm\right)\)
Cho a, b, c là các số thực dương thỏa mãn điều kiện \(\left(a+b-c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\right)=4\)
Chứng minh \(\left(a^4+b^4+c^4\right)\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}\right)\ge2304\)
Theo giả thiết kết hợp sử dụng BĐT AM - GM có:
\(\left(a+b-c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\right)=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+1-\left[c\left(a+b\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right]\)
\(\le\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+1-2\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}=\left[\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1\right]^2\)
Suy ra \(\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1\ge2\Leftrightarrow\sqrt{\dfrac{a}{b}+\dfrac{b}{a}+2}\ge3\)
\(\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge7\)
Khi đó, sử dụng BĐT Cauchy - Schwarz ta có:
\(\left(a^4+b^4+c^4\right)\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}\right)\ge\left[\sqrt{\left(a^4+b^4\right)\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}\right)}+1\right]^2\)
\(=\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+1\right)^2=\left[\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-1\right]^2\ge\left(7^2-1\right)^2=2304\)
Đẳng thức xảy ra khi và chỉ khi \(ab=c^2\) và \(\dfrac{a}{b}+\dfrac{b}{a}=7\)
(a+b-c)(1/a+1/b-c)=(a+b)(1/a+1/b)+1-[c(a+b)+c(1/a+1/b)]<=(a+b)(1/a+1/b)+1-2căn (a+b)(1/a+1/b)
=[(căn (a+b)(1/a+1/b))-1]^2
=>\(\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1>=2\)
=>\(\sqrt{\dfrac{a}{b}+\dfrac{b}{a}+2}>=3\)
=>a/b+b/a>=7
(a^4+b^4+c^4)(1/a^4+1/b^4+1/c^4)>=[căn ((a^4+b^4)(1/a^4+1/b^4))+1]^2
=(a^2/b^2+b^2/a^2+1)^2=[(a/b+b/a)^2-1]^2>=(7^2-1)^2=2304
=>ĐPCM
Cho a,b,c là các số dương. Cm:
a. \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
b. \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
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)
\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Cho a, b, c là các số dương biết abc = 1. Chứng minh rằng: \(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}\ge\dfrac{1}{2}\)
\(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b+1}{12}+\dfrac{c+2}{18}\ge3\sqrt[3]{\dfrac{a^3\left(b+1\right)\left(c+2\right)}{216\left(b+1\right)\left(c+2\right)}}=\dfrac{a}{2}\)
Tương tự: \(\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c+1}{12}+\dfrac{a+2}{18}\ge\dfrac{b}{2}\)
\(\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}+\dfrac{a+1}{12}+\dfrac{b+2}{18}\ge\dfrac{c}{2}\)
Cộng vế:
\(VT+\dfrac{5}{36}\left(a+b+c\right)+\dfrac{7}{12}\ge\dfrac{1}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{13}{36}\left(a+b+c\right)-\dfrac{7}{12}\ge\dfrac{13}{36}.3\sqrt[3]{abc}-\dfrac{7}{12}=\dfrac{1}{2}\) (đpcm)
cho a,b,c là các số thực dương. Chứng minh rằng :
\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(a+b+c\right)\)
AD bđt AM-GM cho 3 số
\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+C}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c}{a^3\left(b+c\right)}.\dfrac{\left(b+c\right)}{4bc}.\dfrac{1}{2b}}=\dfrac{3}{2a}\)
\(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}\ge\dfrac{3}{2a}-\dfrac{3}{4b}-\dfrac{1}{4c}\)
thiết lập bđt tương tự r cộng lại \(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\left(\dfrac{3}{2}-\dfrac{3}{4}-\dfrac{1}{4}\right)\left(a+b+c\right)=\dfrac{1}{2}\left(a+b+c\right)\)
1. cho a,b,c là các số dương .Cmr :
\(\dfrac{a^3+b^3}{ab}+\dfrac{b^3+c^3}{bc}+\dfrac{a^3+c^3}{ac}\ge2\left(a+b+c\right)\)
Chứng minh: \(x^3+y^3\ge xy\left(x+y\right)\left(1\right)\)
\(x^3+y^3\ge xy\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)\ge xy\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^3\ge4xy\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\) đúng
\(\Rightarrow\left(1\right)\) đúng
Áp dụng BĐT \(x^3+y^3\ge xy\left(x+y\right)\)
\(\dfrac{a^3+b^3}{ab}+\dfrac{b^3+c^3}{bc}+\dfrac{c^3+a^3}{ca}\)
\(\ge\dfrac{ab\left(a+b\right)}{ab}+\dfrac{bc\left(b+c\right)}{bc}+\dfrac{ca\left(c+a\right)}{ca}\)
\(=2\left(a+b+c\right)\)
cho các số dương a và b thỏa mãn a+b=1. chứng minh rằng \(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\ge9\)
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 )
Helppppppppppppppppppp
Cho a,b,c là các số thực dương thoả mãn a +b + c <1 . Chứng minh rằng \(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+\left(a+b\right)}+\dfrac{1}{bc+\left(b+c\right)}+\dfrac{1}{ca+\left(c+a\right)}< \dfrac{87}{2}\)
Bất đẳng thức sai, chẳng hạn với \(a=b=10^{-4};c=0,5-a-b\).