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a, Áp dụng \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)
Áp dụng \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\forall x,y>0\)
Ta có: \(A=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2\ge\frac{\left(2+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\ge\frac{\left(2+\frac{4}{a+b}\right)^2}{2}\ge\frac{\left(2+4\right)^2}{2}=18\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
b, Áp dụng \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)
Áp dụng \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\forall x,y,z>0\)
Ta có: \(B=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2+\left(1+\frac{1}{c}\right)^2\ge\frac{\left(3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\ge\frac{\left(3+\frac{9}{a+b+c}\right)^2}{3}\ge\frac{\left(3+6\right)^2}{3}=27\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{2}\)
* Các BĐT phụ bạn tự CM nha! Chúc bạn học tốt
Bài làm:
Bài 1:
Ta có: \(T=8x^2-4x+\frac{1}{4x^2}+15\)
\(=\left(4x^2-4x+1\right)+\left(4x^2+\frac{1}{4x^2}\right)+14\)
\(=\left(2x-1\right)^2+\left(4x^2+\frac{1}{4x^2}\right)+14\)\(\ge0+2\sqrt{4x^2.\frac{1}{4x^2}}+14=2+14=16\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(2x-1\right)^2=0\\4x^2=\frac{1}{4x^2}\end{cases}\Rightarrow x=\frac{1}{2}}\)
Vậy \(Min\left(T\right)=16\)khi \(x=\frac{1}{2}\)
Bài 2:
Ta có: \(ab+bc+ca=3abc\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=3\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\left(1\right)\)
Ta xét \(\frac{a^2}{c\left(c^2+a^2\right)}=\frac{\left(c^2+a^2\right)-c^2}{c\left(c^2+a^2\right)}=\frac{1}{c}-\frac{c}{c^2+a^2}=\frac{1}{c}-\frac{1}{a}.\frac{ac}{c^2+a^2}\ge\frac{1}{c}-\frac{1}{a}.\frac{ac}{2ac}=\frac{1}{c}-\frac{1}{2}a\)
Tương tự ta chứng minh được: \(\frac{b^2}{a\left(a^2+b^2\right)}\ge\frac{1}{a}-\frac{1}{2}b\)và \(\frac{c^2}{b\left(b^2+c^2\right)}\ge\frac{1}{b}-\frac{1}{2}c\)
Cộng vế 3 bất đẳng thức trên lại ta được:
\(P\ge\frac{1}{c}-\frac{1}{2}a+\frac{1}{a}-\frac{1}{2}b+\frac{1}{b}-\frac{1}{2}c\)\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{2}.3=\frac{3}{2}\left(theo\left(1\right)\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}a^2=b^2\\b^2=c^2\\c^2=a^2\end{cases}\Rightarrow a=b=c=1}\)
Vậy \(Min\left(P\right)=\frac{3}{2}\)khi \(a=b=c=1\)
Học tốt!!!!
a.
\(A=\frac{1}{ab}+\frac{1}{a^2+b^2}=\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{1}{2ab}\)
\(\ge\frac{4}{a^2+2ab+b^2}+\frac{1}{2ab}\ge\frac{4}{\left(a+b\right)^2}+\frac{1}{\frac{\left(a+b\right)^2}{2}}=6\)
Dấu "=" khi \(a=b=\frac{1}{2}\)
b.
\(B=\frac{2}{ab}+\frac{3}{a^2+b^2}=3\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{1}{2ab}\)
\(\ge3\cdot\frac{4}{\left(a+b\right)^2}+\frac{1}{\frac{\left(a+b\right)^2}{2}}=14\)
Dấu "=" khi \(a=b=\frac{1}{2}\)
c.
Ta có:
\(x^2+y^2\ge2xy\)
\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)
\(\Leftrightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\) với mọi x,y
Áp dụng ta có:
\(C=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\ge\frac{\left(a+b+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\ge\frac{\left(1+\frac{4}{a+b}\right)^2}{2}=\frac{25}{2}\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
2.
Áp dụng bất đẳng thức Bunhiacopxki ta có:
\(\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2\right]\left[\left(\frac{a}{\sqrt{x}}\right)^2+\left(\frac{b}{\sqrt{y}}\right)^2\right]\ge\left(\sqrt{x}\cdot\frac{a}{\sqrt{x}}+\sqrt{y}\cdot\frac{b}{\sqrt{y}}\right)^2\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{a^2}{x}+\frac{b^2}{y}\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)
Áp dụng nó ta chứng minh được:
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b\right)^2}{x+y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
Áp dụng vào bài làm:
\(D=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ca}+\frac{b^2}{bc+ab}+\frac{c^2}{ca+bc}\)
\(\ge\frac{\left(a+b+c\right)^2}{ab+ca+bc+ab+ca+bc}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
1a
\(A=\frac{3}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^4+b^4}{2}\ge\frac{6}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^2+b^2\right)^2}{2}}{2}\)
\(\ge10+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{4}=10+\frac{1}{16}=\frac{161}{16}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(A_{min}=\frac{161}{16}\)
1b.\(B=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^8+b^8}{4}\ge\frac{2}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^4+b^4\right)^2}{2}}{4}\)
\(\ge6+\frac{\left[\frac{\left(a^2+b^2\right)^2}{2}\right]^2}{8}\ge6+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{32}=6+\frac{1}{128}=\frac{769}{128}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(B_{min}=\frac{769}{128}\)khi \(a=b=\frac{1}{2}\)
3. A có 2n+1 số hạng chia thành n cặp thì thừa 1 số
A= 1/(n+1) + 1/(n+2)...+1/2n+1/(2n+1)+ 1/3n+...+ 1/(3n+1)
Mỗi cặp =1/(2n+1-k)+1/(2n+1+k)=(4n+2)/((2n+1)2-k2) >(4n+2)/(2n+1)2=2/(2n+1)
=>A>(2/(2n+1)).n+1/(2n+1)=1
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
a) \(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\) (ĐKXĐ: \(x\ne\pm1\) )
\(=\left(\frac{x+1+2\left(1-x\right)-5+x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{x+1+2-2x-5+x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{-2}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\frac{2}{x^2-1}.\frac{x^2-1}{1-2x}=\frac{2}{1-2x}\)
b) Để x nhận giá trị nguyên <=> 2 chia hết cho 1 - 2x
<=> 1-2x thuộc Ư(2) = {1;2;-1;-2}
Nếu 1-2x = 1 thì 2x = 0 => x= 0
Nếu 1-2x = 2 thì 2x = -1 => x = -1/2
Nếu 1-2x = -1 thì 2x = 2 => x =1
Nếu 1-2x = -2 thì 2x = 3 => x = 3/2
Vậy ....
Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)
Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)
Ta có:
\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)
Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)
\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)
Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Ta có:\(a+b+c=0\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
Bài 1:
\(A=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=9\)
Đẳng thức xảy ra khi a =b=c=1/3
Bài 2:Buồn ngủ rồi, chắc để đó cho anh Lâm.
Câu 2 có cho a; b dương ko? Nếu cho dương thì đỡ phải xét thêm 1 trường hợp, còn ko cho gì thì xét 2 trường hợp hơi dài