a,b,c >0
a+b+c=3
cmr \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\frac{3}{2}\)
Cho a,b,c > 0.Chứng minh rằng
a,\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)\(\ge\)\(\frac{2}{a+b}\)+\(\frac{2}{b+c}\)+\(\frac{2}{c+a}\)
b,\(\frac{4}{a}\)+\(\frac{5}{b}\)+\(\frac{3}{c}\)\(\ge\)\(4\left(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{c+a}\right)\)
Ta chứng minh BĐT sau với các số dương:
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Thật vậy, BĐT tương đương: \(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Áp dụng:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ; \(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) ; \(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)
Cộng vế với vế:
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
b.
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\Rightarrow\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)
\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)
Cộng vế với vế (1); (2) và (3):
\(\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Cho a,b,c > 0.Chứng minh rằng
a,\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)\(\ge\)\(\frac{2}{a+b}\)+\(\frac{2}{b+c}\)+\(\frac{2}{c+a}\)
b,\(\frac{4}{a}\)+\(\frac{5}{b}\)+\(\frac{3}{c}\)\(\ge\)\(4\left(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{c+a}\right)\)
Cho a, b, c, d dương. CM:
1) \(\frac{a^2}{b^5}+\frac{b^2}{c^5}+\frac{c^2}{d^5}+\frac{d^2}{a^5}\ge\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)
2) \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{a+b+c}{\sqrt[3]{abc}}\)
3) \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{d^2}+\frac{d^2}{a^2}\ge\frac{a+b+c+d}{\sqrt[4]{abcd}}\)
4) \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge9;a+b+c\le1\)
Làm tạm một câu rồi đi chơi, lát làm cho.
4)
Áp dụng bất đẳng thức Cauchy-Schwarz :
\(VT\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1}=9\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
2/ Cô: \(\frac{2a}{b}+\frac{b}{c}\ge3\sqrt[3]{\frac{a.a.b}{b.b.c}}=3\sqrt[3]{\frac{a^3}{abc}}=\frac{3a}{\sqrt[3]{abc}}\)
Tương tự hai BĐT còn lại và cộng theo vế thu được:
\(3.VT\ge3.VP\Rightarrow VT\ge VP^{\left(Đpcm\right)}\)
Đẳng thức xảy ra khi a = b= c
cho a, b, c>0. CMR a\(\frac{a^3}{b}\ge a^2+ab-b^2\)
CM \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\)
Cho a, b, c là độ dài 3 cạnh của tam giác CM \(\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Tự nhiên lục được cái này :'(
3. Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :
\(\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{\left(1+1\right)^2}{a+b-c+b+c-a}=\frac{4}{2b}=\frac{2}{b}\)
\(\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{\left(1+1\right)^2}{b+c-a+c+a-b}=\frac{4}{2c}=\frac{2}{c}\)
\(\frac{1}{a+b-c}+\frac{1}{c+a-b}\ge\frac{\left(1+1\right)^2}{a+b-c+c+a-b}=\frac{4}{2a}=\frac{2}{a}\)
Cộng theo vế ta có điều phải chứng minh
Đẳng thức xảy ra <=> a = b = c
Cho a,b,c dương và a+b+c=3. c/m\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge\frac{3}{2}\)
và \(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\ge\frac{3}{2}\)
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge\frac{3}{2}\)
+) cm: \(\frac{1}{a^2+1}=1-\frac{a^2}{a^2+1}\ge1-\frac{a^2}{2a}=1-\frac{a}{2}\)
\(\frac{1}{b^2+1}\ge1-\frac{b}{2}\)
\(\frac{1}{c^2+1}\ge1-\frac{c}{2}\)
Cộng theo vế:
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge3-\frac{a+b+c}{2}=\frac{3}{2}\)
Dấu "=" xảy ra <=> a = b = c = 1
Cho a,b,c >0 thỏa mãn a+b+c\(\le\)\(\frac{3}{2}\).Chứng minh
a,\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)\(\ge\)6
b,a+ b+ c+ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)\(\ge\)\(\frac{15}{2}\)
a)Áp dụng BĐT cosi-schwart:
`A=1/a+1/b+1/c>=9/(a+b+c)`
Mà `a+b+c<=3/2`
`=>A>=9:3/2=6`
Dấu "=" `<=>a=b=c=1/2`
b)Áp dụng BĐT cosi:
`a+1/(4a)>=1`
`b+1/(4b)>=1`
`c+1/(4c)>=1`
`=>a+b+c+1/(4a)+1/(4b)+1/(4c)>=3`
Ta có:
`1/a+1/b+1/c>=6`(Ở câu a)
`=>3/4(1/a+1/b+1/c)>=9/2`
`=>a+b+c+1/(a)+1/(b)+1/(c)>=3+9/2=15/2`
Dấu "=" `<=>a=b=c=1/2`
a)Áp dụng BĐT cosi-schwart:
A=1a+1b+1c≥9a+b+cA=1a+1b+1c≥9a+b+c
Mà a+b+c≤32a+b+c≤32
⇒A≥9:32=6⇒A≥9:32=6
Dấu "=" ⇔a=b=c=12⇔a=b=c=12
b)Áp dụng BĐT cosi:
a+14a≥1a+14a≥1
b+14b≥1b+14b≥1
c+14c≥1c+14c≥1
⇒a+b+c+14a+14b+14c≥3⇒a+b+c+14a+14b+14c≥3
Ta có:
1a+1b+1c≥61a+1b+1c≥6(Ở câu a)
⇒34(1a+1b+1c)≥92⇒34(1a+1b+1c)≥92
⇒a+b+c+1a+1b+1c≥3+92=152⇒a+b+c+1a+1b+1c≥3+92=152
Dấu "=" ⇔a=b=c=12
Cho a,b,c>0 và a+b+c=3CMR
\(\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ac}\ge\frac{3}{2}\)
@Ace Legona
Áp dụng BĐT AM-GM ta có:
\(VT=\dfrac{1}{a}-\dfrac{a}{c+a^2}+\dfrac{1}{b}-\dfrac{b}{a+b^2}+\dfrac{1}{c}-\dfrac{c}{b+c^2}\)
\(=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\left(\dfrac{a}{c+a^2}+\dfrac{b}{a+b^2}+\dfrac{c}{b+c^2}\right)\)
\(\ge\dfrac{9}{a+b+c}-\left(\dfrac{a}{2a\sqrt{c}}+\dfrac{b}{2b\sqrt{a}}+\dfrac{c}{2c\sqrt{b}}\right)\)
\(\ge3-\left(\dfrac{1}{2\sqrt{c}}+\dfrac{1}{2\sqrt{a}}+\dfrac{1}{2\sqrt{b}}\right)\)\(=3-\left(\dfrac{2\sqrt{a}}{4a}+\dfrac{2\sqrt{b}}{4b}+\dfrac{2\sqrt{c}}{4c}\right)\)
\(\ge3-\left(\dfrac{a+1}{4a}+\dfrac{b+1}{4b}+\dfrac{c+1}{4c}\right)\)
\(=3-\left(\dfrac{3}{4}+\dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{4c}\right)\ge3-\left(\dfrac{3}{4}+\dfrac{9}{4\left(a+b+c\right)}\right)=\dfrac{3}{2}\)
Khi \(a=b=c=1\)
CMR:
\(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\ge\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\)
\(\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{a^3}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\)
\(\)
Cho a,b,c > 0. CMR:
1. \(a^3+b^3+c^3\ge3abc\)
2. \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}\)
3. \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\)
4. \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
5. \(\frac{1}{\left(a+1\right)^2}+\frac{1}{\left(b+1\right)^2}\ge\frac{1}{ab+1}\)
6.\(\frac{1}{1+a^3}+\frac{1}{1+b^3}+\frac{1}{1+c^3}\ge\frac{3}{1+abc}\)
1) Cho a, b, c > 0. Chứng minh: \(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
2) Cho \(a,b,c\in R\).
a) Chứng minh: \(\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right)\ge4\left(a+b+c+1\right)^2\)
b) Chứng minh: \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{16}\left(a+b+c+1\right)^2\)
3) Cho \(a,b,c\in R\)Chứng minh: \(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
2) Theo nguyên lí Dirichlet, trong ba số \(a^2-1;b^2-1;c^2-1\) có ít nhất hai số nằm cùng phía với 1.
Giả sử đó là a2 - 1 và b2 - 1. Khi đó \(\left(a^2-1\right)\left(b^2-1\right)\ge0\Leftrightarrow a^2b^2-a^2-b^2+1\ge0\)
\(\Rightarrow a^2b^2+3a^2+3b^2+9\ge4a^2+4b^2+8\)
\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\ge4\left(a^2+b^2+2\right)\)
\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right)\ge4\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\) (2)
Mà \(4\left[\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\right]\ge4\left(a+b+c+1\right)^2\) (3)(Áp dụng Bunhicopxki và cái ngoặc vuông)
Từ (2) và (3) ta có đpcm.
Sai thì chịu
Xí quên bài 2 b:v
b) Không mất tính tổng quát, giả sử \(\left(a^2-\frac{1}{4}\right)\left(b^2-\frac{1}{4}\right)\ge0\)
Suy ra \(a^2b^2-\frac{1}{4}a^2-\frac{1}{4}b^2+\frac{1}{16}\ge0\)
\(\Rightarrow a^2b^2+a^2+b^2+1\ge\frac{5}{4}a^2+\frac{5}{4}b^2+\frac{15}{16}\)
Hay \(\left(a^2+1\right)\left(b^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{3}{4}\right)\)
Suy ra \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{1}{4}+\frac{1}{2}\right)\left(\frac{1}{4}+\frac{1}{4}+c^2+\frac{1}{2}\right)\)
\(\ge\frac{5}{4}\left(\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c+\frac{1}{2}\right)^2=\frac{5}{16}\left(a+b+c+1\right)^2\) (Bunhiacopxki) (đpcm)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)
Cách nữa cho bài 2:
2a) Ta có: \(4\left(a^2+1+2\right)\left(1+1+\frac{\left(b+c\right)^2}{2}\right)\ge4\left(a+b+c+1\right)^2\)
Hay \(4\left(a^2+3\right)\left(2+\frac{\left(b+c\right)^2}{2}\right)\ge4\left(a+b+c+1\right)^2=VP\)
Như vậy ta quy bài toán về chứng minh: \(\left(b^2+3\right)\left(c^2+3\right)\ge4\left(2+\frac{\left(b+c\right)^2}{2}\right)\)
\(\Leftrightarrow b^2c^2+b^2+c^2+1\ge4bc\Leftrightarrow\left(bc-1\right)^2+\left(b-c\right)^2\ge0\)(đúng)
Đẳng thức xảy ra khi a = b = c = 1
b) Áp dụng BĐT Bunhiacopxki:\(\left(a^2+\frac{1}{4}+\frac{1}{4}+\frac{1}{2}\right)\left(\frac{1}{4}+b^2+c^2+\frac{1}{2}\right)\ge\frac{1}{4}\left(a+b+c+1\right)^2\)
\(\Rightarrow\frac{5}{4}\left(a^2+1\right)\left(b^2+c^2+\frac{3}{4}\right)\ge\frac{5}{16}\left(a+b+c+1\right)^2\)
Từ đó ta có thể quy bài toán về chứng minh: \(\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{4}\left(b^2+c^2+\frac{3}{4}\right)\)
...
Bài 3:Sửa đề a, b, c >0
Có: \(\frac{a^3}{b^2}+\frac{a^3}{b^2}+b\ge3\sqrt[3]{\frac{a^6}{b^3}}=\frac{3a^2}{b}\)
Tương tự: \(\frac{2b^3}{c^2}+c\ge\frac{3b^2}{c};\frac{2c^3}{a^2}+a\ge\frac{3c^2}{a}\)
Cộng theo vế 3 BĐT trên: \(2\left(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\right)+a+b+c\ge3\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\)
\(=2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\)
\(\ge2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+a+b+c\)
Từ đó ta có đpcm.