\(\frac{a}{b}\left(b>0\right)\frac{a+n}{b+n}\)
Chứng minh:
\(a^n+b^n+c^n\ge\left(\frac{a+2b}{3}\right)^n+\left(\frac{b+2c}{3}\right)^n+\left(\frac{c+2a}{3}\right)^n,\forall a,b,c>0;n\in N\)
Cho 3 số thực a,b,c thỏa mãn \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\) = 0. CMR
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}\) = 0
#Xin lỗi cho t gửi nhờ
Ta có:
\(\frac{x}{1+4y^2}=\frac{x\left(1+4y^2\right)-4xy^2}{1+4y^2}=x-\frac{4xy^2}{1+4y^2}\)
Áp dụng BĐT AM-GM cho các số không âm, ta có:
\(1+4y^2\ge4y\Rightarrow\frac{4xy^2}{1+4y^2}\le\frac{4xy^2}{4y}=xy\) \(\Rightarrow-\frac{4xy^2}{1+4y^2}\ge-xy\)
\(\Rightarrow\frac{x}{1+4y^2}\ge x-xy\)
Tương tự ta có: \(P\ge x+y+z-\left(xy+yz+zx\right)\)
C/m ở câu a: \(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\)
\(\Rightarrow-\left(xy+yz+zx\right)\ge-\frac{\left(x+y+z\right)^2}{3}\)
Khi đó: \(P\ge x+y+z-\frac{\left(x+y+z\right)^2}{3}=\frac{3}{2}-\frac{\left(\frac{3}{2}\right)^2}{3}=\frac{3}{4}\)
Dấu ''='' xảy ra \(\Leftrightarrow x=y=z=\frac{1}{2}\)
Vậy...
Lời giải:
Nên bổ sung thêm điều kiện $a,b,c$ đôi một phân biệt. Đặt biểu thức cần chứng minh bằng $0$ là $P$
Ta có:
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow \left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)=0\)
\(\Leftrightarrow \frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}+\frac{b}{(b-c)(c-a)}+\frac{c}{(b-c)(a-b)}+\frac{a}{(c-a)(b-c)}+\frac{c}{(c-a)(a-b)}+\frac{a}{(a-b)(b-c)}+\frac{b}{(a-b)(c-a)}=0\)
\(\Leftrightarrow P+\frac{b(a-b)+c(c-a)+a(a-b)+c(b-c)+a(c-a)+b(b-c)}{(a-b)(b-c)(c-a)}=0\)
\(\Leftrightarrow P+\frac{0}{(a-b)(b-c)(c-a)}=0\Rightarrow P=0\) (đpcm)
Cho a, b, c khác 0 thỏa mãn a+b+c=0. Tính \(A=\left(1+\frac{a}{b}\right).\left(1+\frac{b}{c}\right).\left(1+\frac{c}{a}\right)\)
Cho a,b,c #0 giải phân thức:\(\frac{x}{\left(a-b\right).\left(a-c\right)}+\frac{x}{\left(b-a\right).\left(b-c\right)}+\frac{x}{\left(c-a\right).\left(c-b\right)}=2\)
Với \(a,b,c\ne0\) ta có :
\(\frac{x}{\left(a-b\right)\left(a-c\right)}+\frac{x}{\left(b-a\right)\left(b-c\right)}+\frac{x}{\left(c-a\right)\left(c-b\right)}=2\)
\(\Leftrightarrow x\left(\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\right)=2\)
\(\Leftrightarrow x\left(\frac{1}{a-b}-\frac{1}{a-c}+\frac{1}{b-a}-\frac{1}{b-c}+\frac{1}{c-a}-\frac{1}{c-b}\right)=2\)
\(\Leftrightarrow0x=2\) ( vô lí)
Vậy...
cho a,b > 0 thỏa mãn a + b = 2. Cmr: \(2\left(a^2+b^2\right)-6\left(\frac{a}{b}+\frac{b}{a}\right)+9\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge10\)
\(2\left(a^2+b^2\right)-6\left(\frac{a}{b}+\frac{b}{a}\right)+9\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\\ =\left(\frac{3}{a^2}+3b^2\right)+\left(\frac{3}{b^2}+3a^2\right)-\left(a^2+2ab+b^2\right)-6\left(\frac{a}{b}+\frac{b}{a}\right)+6\left(ab+ab+\frac{1}{a^2}+\frac{1}{b^2}\right)-10ab\)
Áp dụng bất đẳng thức Cô-si với 2 số không âm:
\(\Rightarrow2\left(a^2+b^2\right)-6\left(\frac{a}{b}+\frac{b}{a}\right)+9\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\\ \ge2\sqrt{\frac{3}{a^2}\cdot3b^2}+2\sqrt{\frac{3}{b^2}\cdot3a^2}-\left(a+b\right)^2-6\left(\frac{a}{b}+\frac{b}{a}\right)+6\cdot4\sqrt{ab\cdot ab\cdot\frac{1}{a^2}\cdot\frac{1}{b^2}}-\frac{10\left(a+b\right)^2}{4}\\ =\frac{6b}{a}+\frac{6a}{b}-4-6\left(\frac{a}{b}+\frac{b}{a}\right)+24-10\\ =10\)
Dấu "=" xảy ra khi \(a=b=1\)
cho a,b,c > 0 thỏa mãn \(2\left(\frac{a}{b}+\frac{b}{a}\right)+c\left(\frac{a}{b^2}+\frac{b}{a^2}\right)=6\)
Tìm GTNN của \(A=\frac{bc}{a\left(2b+c\right)}+\frac{ac}{b\left(2a+c\right)}+\frac{4ab}{c\left(a+b\right)}\)
Cho a,b>0 thoả mãn a+b=1
CMR: \(\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)\left(c+\frac{1}{c}\right)\ge\frac{25}{4}\)
Ta có: \(\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)\left(c+\frac{1}{c}\right)\)
\(=\left(ab+\frac{1}{ab}+\frac{a}{b}+\frac{b}{a}\right)\left(c+\frac{1}{c}\right)\)
\(=\left[ab+\frac{1}{16ab}+\frac{15}{16ab}+\left(\frac{a}{b}+\frac{b}{a}\right)\right]\left(c+\frac{1}{c}\right)\)
\(\ge\left[2\sqrt{ab.\frac{1}{16ab}}+\frac{15}{4\left(a+b\right)^2}+2\sqrt{\frac{a}{b}.\frac{b}{a}}\right]\left(2\sqrt{c.\frac{1}{c}}\right)\)
\(\ge\frac{25}{2}\left(Đpcm\right)\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=\frac{1}{2};c=1\)
Giúp mình với:
1. Cho 2 số nguyên a và b ( b \(\ne\)0 ). Khẳng định nào dưới đây là đúng ?
A. \(\frac{-\left(-a\right)}{-b}=\frac{-a}{-b}\) B. \(\frac{-a}{-b}=\frac{-a}{-\left(-b\right)}\) C. \(\frac{-\left(-a\right)}{-b}=\frac{a}{b}\) D. \(\frac{-\left(-a\right)}{-\left(-b\right)}=\frac{a}{b}\)
2. Cho 2 phân số bằng nhau \(\frac{a}{b}=\frac{c}{d}\) (a,b,c,d \(\varepsilon\)Z; b,d \(\ne\)0). Chứng minh rằng \(\frac{a\pm b}{_{ }b}=\frac{c\pm d}{d}\)
Bài 1: D
Bài 2:
Ta có: \(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}\pm1=\frac{c}{d}\pm1\)
\(\Rightarrow\frac{a\pm b}{b}=\frac{c\pm d}{d}\)(đpcm)
Cho a,b,c>0 thỏa mãn abc=1. Chứng minh
\(\frac{a}{\left(a+1\right)\left(b+1\right)}+\frac{b}{\left(b+1\right)\left(c+1\right)}+\frac{c}{\left(c+1\right)\left(a+1\right)}\ge\frac{3}{4}\)
Đặt \(a=\frac{x}{y};b=\frac{y}{z};c=\frac{z}{x}\). Xét hiệu 2 vế:
\(VT-VP=\frac{\sum\limits_{cyc} x(y-z)^2}{4(x+y)(y+z)(z+x)} \geq 0\)
Ta có đpcm.