Cho abc = 2 .Rút gọn biểu thức :
A=\(\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)
Cho abc = 2 .Rút gọn biểu thức :
A=\(\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)
Thay abc = 2 vào biểu thức A ta được:
\(A=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{abc\cdot c}{ac+abc+abc}\\ A=\dfrac{1}{b+1+bc}+\dfrac{b}{bc+b+1}+\dfrac{bc}{1+bc+b}\\ A=\dfrac{1+b+bc}{1+b+bc}\\ A=1\)
CMR nếu a, b\(\in N^{ }\)( a,b # 0) và a + b = 1 thì:
\(\left(a+\dfrac{1}{b}\right)^2+\left(b+\dfrac{1}{a}\right)^2\ge\dfrac{25}{2}\)
Các bn júp mk lm bài này.
Mk cảm ơn trước nha!!!
cach khac\(\left(a+\dfrac{1}{b}\right)^2+\left(b+\dfrac{1}{a}\right)^2\ge\dfrac{1}{2}\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2\ge\dfrac{1}{2}\left(a+b+\dfrac{4}{a+b}\right)^2=\dfrac{25}{2}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1^2+1^2\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2=1\)
\(\Rightarrow2\left(a^2+b^2\right)\ge1\Rightarrow a^2+b^2\ge\dfrac{1}{2}\)
Áp dụng BĐT Holder ta có:
\(\left(a+b\right)\left(a+b\right)\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\ge\left(1+1\right)^3=8\)
Lại có:
\(\left(a+\dfrac{1}{b}\right)^2+\left(b+\dfrac{1}{a}\right)^2=4+a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge4+\dfrac{1}{2}+8=\dfrac{25}{2}\)
Nếu x, y, z > 0 thì \(\dfrac{x^3}{y^2}+\dfrac{y^3}{z^2}+\dfrac{z^3}{x^2}\ge x+y+z\)
Áp dụng bđt AM-GM:
\(\dfrac{x^3}{y^2}+y+y\ge3\sqrt[3]{x^3}=3x\)
\(\dfrac{y^3}{z^2}+z+z\ge3\sqrt[3]{y^3}=3y\)
\(\dfrac{z^3}{x^2}+x+x\ge3\sqrt[3]{z^3}=3z\)
Cộng theo vế suy ra: \(\dfrac{x^3}{y^2}+\dfrac{y^3}{z^2}+\dfrac{z^3}{x^2}\ge x+y+z\)
"=" khi a=b=c
Đề bài là cmr nhé
BT1: Cho a+b>1. Chứng minh: a4+b4>=\(\dfrac{1}{8}\)
BT2: Cho a,b,c>0. Chứng minh rằng: \(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}>=a+b+c\)
Bài 1: \(a+b\ge1\). cm \(a^4+b^4\ge\dfrac{1}{8}\)
ta có : \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2=\dfrac{1}{2}\)(BĐT bunyakovsky)
Áp dụng BĐt bunyakovsky 1 lần nữa:
\(a^4+b^4\ge\dfrac{1}{2}\left(a^2+b^2\right)^2\ge\dfrac{1}{2}.\dfrac{1}{4}=\dfrac{1}{8}\)
dấu = xảy ra khi \(a=b=\dfrac{1}{2}\)
Bài 2:
Áp dụng BĐT bunyakovsky dạng đa thức và phân thức:
\(\left(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\right)\left(a+b+c\right)\ge\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)^2\ge\left[\dfrac{\left(a+b+c\right)^2}{a+b+c}\right]^2=\left(a+b+c\right)^2\)
do đó \(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\)
dấu = xảy ra khi a=b=c
Bài 1:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1^2+1^2\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2=1\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge1\Rightarrow a^2+b^2\ge\dfrac{1}{2}\)
Lại theo Cauchy-Schwarz lần nữa:
\(\left[\left(1^2\right)^2+\left(1^2\right)^2\right]\left[\left(a^2\right)^2+\left(b^2\right)^2\right]\ge\left(a^2+b^2\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow2\left(a^4+b^4\right)\ge\dfrac{1}{4}\Leftrightarrow a^4+b^4\ge\dfrac{1}{8}\)
Đẳng thức xảy ra khi \(a=b=\dfrac{1}{2}\)
Bài 2:
Trước tiên ta chứng minh \(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\)
Ta chứng minh bổ đề: \(\dfrac{a^3}{b^2}\ge\dfrac{a^2}{b}+a-b\)
\(\Leftrightarrow a^3+b^3\ge ab\left(a+b\right)\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)(đúng)
Viết các BĐT tương tự và cộng lại
\(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge\dfrac{a^2}{b}+a-b+\dfrac{b^2}{c}+b-c+\dfrac{c^2}{a}+c-a=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\left(1\right)\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\left(2\right)\)
Từ \((1);(2)\) ta thu được ĐPCM
BT2) Áp dụng BĐT AM-GM,ta có:
\(\dfrac{a^3}{b^2}\)+b+b\(\geq\) 3\(\sqrt[3]{\dfrac{a^3}{b^2}.b.b}\) = 3a
\(\dfrac{b^3}{c^2}\)+c+c\(\geq\) 3\(\sqrt[3]{\dfrac{b^3}{c^2}.c.c}\) = 3b
\(\dfrac{c^3}{a^2}\)+a+a\(\geq\) 3\(\sqrt[3]{\dfrac{c^3}{a^2}.a.a}\) = 3a
Cộng các BĐT lại với nhau ta được:
\(\dfrac{a^3}{b^2}\)+\(\dfrac{b^3}{c^2}\)+\(\dfrac{c^3}{a^2}\)+2a+2b+2c\(\geq\) 3a+3b+3c
\(\Leftrightarrow\) \(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\) \(\geq\) a+b+c
Dấu = xảy ra khi a=b=c