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, \(\left(a+b+c\right)\times\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
a,b,c>0.C/m
a, \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)
b, \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
a)Theo bất đẳng thức cauchy:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\Rightarrow\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{4}{a+b}.\left(a+b\right)\)
\(\Rightarrow\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)
Dấu "=" xảy ra khi: \(a=b\)
Ta có điều phải chứng minh
b)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\ge\dfrac{9}{a+b+c}.\left(a+b+c\right)\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\ge9\)
Dấu "=" xảy ra khi:
\(a=b=c\)
Ta có điều phải chứng minh
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: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=3\cdot\sqrt[3]{abc}\cdot3\cdot\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\cdot\sqrt[3]{a\cdot b\cdot c\cdot\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\)
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)
Với a, b, c là những số thực dương thỏa mãn \(\left(a+b\right)\left(b+c\right)\)\(\left(c+a\right)\)=1
Chứng minh rằng \(\dfrac{a}{b\left(b+2c\right)^2}\)+\(\dfrac{b}{c\left(c+2a\right)^2}\)+\(\dfrac{c}{a\left(a+2b\right)^2}\)≥\(\dfrac{4}{3}\)
CM CÁC BẤT ĐẲNG THỨC SAU
A) \(\left(A+B\right)\left(\dfrac{1}{A}+\dfrac{1}{B}\right)\ge4\)
B) \(\left(A+B+C\right)\left(\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\right)\ge9\)
C) \(\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\ge\dfrac{9}{A+B+C}\)
c) Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có :
\(\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\ge\dfrac{\left(1+1+1\right)^2}{A+B+C}=\dfrac{9}{A+B+C}\)
Dấu "=" xảy ra khi và chỉ khi\(\dfrac{1}{A}=\dfrac{1}{B}=\dfrac{1}{C}\)
Cho a,b,c là độ dài ba cạnh của một tam giác .Chứng minh rằng :
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3\left(a-b\right)\left(b-c\right)\left(c-a\right)}{abc}\ge9\)
Cho a, b, c là ba số dương thỏa mãn \(abc\)=1. Chứng minh rằng:
\(\dfrac{1}{a^3\left(b+c\right)}\)+\(\dfrac{1}{b^3\left(a+c\right)}\)+\(\dfrac{1}{c^3\left(a+b\right)}\)≥\(\dfrac{3}{2}\)
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 \(\Leftrightarrow a=b=c=1\)
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 \(a=b=10^{-4};c=0,5-a-b\).
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 )