Cho a, b, c > 0. CMR: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)≥\(\dfrac{3}{a+2b}+\dfrac{3}{b+2c}+\dfrac{3}{c+2a}\)
Cho a, b, c > 0 . CMR :
\(\dfrac{a^3}{\left(2a+b\right)\left(2b+c\right)}+\dfrac{b^3}{\left(2b+c\right)\left(2c+a\right)}+\dfrac{c^3}{\left(2c+a\right)\left(2a+b\right)}\le\dfrac{a+b+c}{9}\)
Dấu >= hay <= vậy bạn? Bạn xem lại đề.
Bài 1: Cho a,b,c là những số dương thỏa mãn: a+b+c=3
CMR: \(\dfrac{a^2}{a+2b^3}+\dfrac{b^2}{b+2c^3}+\dfrac{c^2}{c+2a^3}\ge1\)
Bài 2: Cho a, b, c thỏa mãn: ab+bc+ca=3
CMR: \(\dfrac{a}{2b^3+1}+\dfrac{b}{2c^3+1}+\dfrac{c}{2a^3+1}\ge1\)
Bài 3: Cho a, b, c > 0. CMR: \(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+3b\)
Dấu = xảy ra khi a=b=2c
Cho a,b,c>0. CMR
\(\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\le\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\)
\(\dfrac{1}{a+3b}+\dfrac{1}{a+b+2c}\ge\dfrac{4}{2a+4b+2c}=\dfrac{2}{a+2b+c}\)
Tương tự: \(\dfrac{1}{b+3c}+\dfrac{1}{b+c+2a}\ge\dfrac{2}{a+b+2c}\)
\(\dfrac{1}{c+3a}+\dfrac{1}{a+c+2b}\ge\dfrac{2}{2a+b+c}\)
Cộng vế với vế và rút gọn:
\(\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\)
Dấu "=" xảy ra khi \(a=b=c\)
Cho a, b, c > 0, a + b + c = 1
CMR : \(\dfrac{a}{2a+b+c}+\dfrac{b}{a+2b+c}+\dfrac{c}{a+b+2c}\le\dfrac{3}{4}\)
Do \(a+b+c=1\) nên Bất đẳng thức trên tương đương:
\(\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}\le\dfrac{3}{4}\)
\(\Leftrightarrow\left(1-\dfrac{1}{1+a}\right)+\left(1-\dfrac{1}{1+b}\right)+\left(1-\dfrac{1}{1+c}\right)\le\dfrac{3}{4}\)
\(\Leftrightarrow3-\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)\le\dfrac{3}{4}\)
Áp dụng BĐT cauchy-schwarz engel với a;b;c>0 ta có:
\(3-\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)\le3-\dfrac{\left(1+1+1\right)^2}{1+a+1+b+1+c}=3-\dfrac{9}{4}=\dfrac{3}{4}\)
Ta có:
\(\dfrac{a}{2a+b+c}+\dfrac{b}{a+2b+c}+\dfrac{c}{a+b+2c}=\dfrac{a}{\left(a+b\right)+\left(a+c\right)}+\dfrac{b}{\left(a+b\right)+\left(b+c\right)}+\dfrac{c}{\left(a+c\right)+\left(b+c\right)}=\dfrac{a}{4}.\dfrac{4}{\left(a+b\right)+\left(a+c\right)}+\dfrac{b}{4}.\dfrac{4}{\left(a+b\right)+\left(b+c\right)}+\dfrac{c}{4}.\dfrac{4}{\left(a+c\right)+\left(b+c\right)}=\dfrac{a}{4}.\dfrac{\left(1+1\right)^2}{\left(a+b\right)+\left(a+c\right)}+\dfrac{b}{4}.\dfrac{\left(1+1\right)^2}{\left(a+b\right)+\left(b+c\right)}+\dfrac{c}{4}.\dfrac{\left(1+1\right)^2}{\left(a+c\right)+\left(b+c\right)}\)Áp dụng BĐT Cauchy - Schwarz:
\(VT\le\dfrac{a}{4}.\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{b}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)+\dfrac{c}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)=\dfrac{1}{4}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)=\dfrac{1}{4}.3=\dfrac{3}{4}\)\("="\Leftrightarrow a=b=c=\dfrac{1}{3}\)
Cho a,b,c>0 và abc=1. CMR
A = \(\dfrac{a}{2a^3+1}+\dfrac{b}{2b^3+1}+\dfrac{c}{2c^3+1}\le1\)
Cho a,b,c > 0:abc=1
Cmr: \(\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2}\)
Ta có:
\(a^2+b^2\ge2ab\)
\(b^2+1\ge2ab\)
\(\Rightarrow a^2+2ab^2+3\ge2\left(ab+b+1\right)\)
\(\Rightarrow\dfrac{1}{a^2+2b^2+3}< \dfrac{1}{2.\left(ab+b+1\right)}\)
Tương tự:
\(\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2}.\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\right)\)
Mặt khác:
\(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}=\dfrac{1}{ab+b+1}+\dfrac{ab}{ab^2c+abc+ab}+\dfrac{b}{bca+ab+b}=1\)
\(\Rightarrow\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2}\)
\(\Leftrightarrow a=b=c=1\)
\(\Rightarrow\) Đpcm.
Áp dụng BĐT AM - GM, ta có:
\(a^2+2b^2+3\)
\(=\left(a^2+b^2\right)+\left(b^2+1\right)+2\)
\(\ge2ab+2b+2\)
Tương tự, ta có: \(b^2+2c^2+3\ge2bc+2c+2\) và \(c^2+2a^2+3\ge2ac+2a+2\)
\(VT=\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\)
\(\le\dfrac{1}{2ab+2b+2}+\dfrac{1}{2bc+2c+2}+\dfrac{1}{2ac+2a+2}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ac+a+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{abc}{bc+c+abc}+\dfrac{abc}{ac+a^2bc+abc}\right)\) (Thay abc = 1)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{ab}{b+1+ab}+\dfrac{b}{1+ab+b}\right)\)
\(=\dfrac{1}{2}\times\dfrac{1+ab+b}{ab+b+1}\)
\(=\dfrac{1}{2}=VP\left(\text{đ}pcm\right)\)
Dấu "=" xảy ra khi a = b = c = 1
Cho 3 số dương a,b,c. CMR: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)
Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\) ta được
\(\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{2b}\ge\dfrac{9}{2\left(a+2b\right)}\)
\(\dfrac{1}{2b}+\dfrac{1}{2c}+\dfrac{1}{2c}\ge\dfrac{9}{2\left(b+2c\right)}\)
\(\dfrac{1}{2c}+\dfrac{1}{2a}+\dfrac{1}{2a}\ge\dfrac{9}{2\left(c+2a\right)}\)
Cộng các BĐT theo vế
\(\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{9}{2}\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)
Dấu " = " xảy ra khi a = b = c ( a,b,c > 0 )
Cho a;b;c>0 tm a+b+c=3
CMR \(\dfrac{2b+c}{a}+\dfrac{2c+a}{b}+\dfrac{2a+b}{c}+\dfrac{18abc}{ab+bc+ac}\ge12\)
Bài 1: CMR:
\(a,\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(b,\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}\ge1\) với a+b+c=3
Bài 2: \(a,b,c\in N,a+b+c=2021\)
Tìm GTNN \(P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Bài 1:
a) Áp dụng bđt Cô - si:
\(\dfrac{a}{b^2}+\dfrac{1}{a}\ge\dfrac{2}{b}\)
Tương tự với 2 phân thức còn lại của vế trái rồi cộng lại, ta có:
\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\)
=> đpcm
Bài dù a + b + c = 2021 hay 1 số bất kì thì bđt luôn \(\ge\dfrac{3}{2}\). Bạn có thể tham khảo bđt Nesbitt
Bài 2:
\(P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(=\dfrac{2021-\left(b+c\right)}{b+c}+\dfrac{2021-\left(c+a\right)}{c+a}+\dfrac{2021-\left(a+b\right)}{a+b}\)
\(=2021\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)-3\)
Áp dụng BĐT Svacxo, ta có
\(P\) ≥ \(\dfrac{9}{2}-3=\dfrac{3}{2}\)
Dấu"=" ⇔ ...
Sau khi đã đi tham khảo 7749 người thì đã cho ra một kết quả:v
Bài 2. \(P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(P=\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1+\dfrac{c}{a+b}+1-3\)
\(P=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)
\(P=\dfrac{(2a+2b+3c)( \dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b})}{2}-3 ≥ \dfrac{9}{2}-3=\dfrac{3}{2}\)
Dấu `"="` xảy ra:
\(\Leftrightarrow \begin{cases} a=b=c\\ a+b+c=2021 \end{cases} \)
\(\Leftrightarrow a=b=c=\dfrac{2021}{3}\)
Vậy \(min \) \(P=\dfrac{3}{2}\) khi \(a=b=c=\dfrac{2021}{3}\)