Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng bđt AM - GM ta có :
\(\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{a^2}}\ge\sqrt{2\frac{a^2}{b^2}}+\sqrt{2\frac{b^2}{a^2}}=\sqrt{2}\frac{a}{b}+\sqrt{2}\frac{b}{a}\)
\(=\sqrt{2}\left(\frac{a}{b}+\frac{b}{a}\right)\ge\sqrt{2}.2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\sqrt{2}\)
a) Áp dụng BĐT Cauchy cho 2 số không âm , ta có:
\(\dfrac{a+b}{2}\ge\sqrt{ab}\)
\(\Rightarrow a+b\ge2\sqrt{ab}\)
b) Xét hiệu:
\(\dfrac{a}{b}+\dfrac{b}{a}-2=\dfrac{a^2+b^2-2ab}{ab}=\dfrac{\left(a-b\right)^2}{ab}\ge0\) ( luôn đúng)
=> \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
a) a + b ≥ 2\(\sqrt{ab}\) ( a > 0 ; b > 0 )
⇔ a - 2\(\sqrt{ab}\) + b ≥ 0
⇔ \(\left(\sqrt{a}-\sqrt{b}\right)^2\) ≥ 0 ( luôn đúng )
b) Áp dụng BĐT Cô-si :
x2 + y2 ≥ 2xy ( x > 0 ; y > 0)
⇒ a2 + b2 ≥ 2ab ( a > 0 ; b > 0)
⇔ \(\dfrac{a^2+b^2}{ab}\) ≥ 2
⇔\(\dfrac{a}{b}+\dfrac{b}{a}\) ≥ 2
a) Ta có: \(\left(a-b\right)^2\ge0\)
=>\(a^2+b^2-2ab\ge0\left(đpcm\right)\)
b) \(\left(a+b\right)^2\ge0\)
=> \(a^2+b^2+2ab\ge0\)
<=> \(a^2+b^2\ge-2ab\)
<=> \(\dfrac{a^2+b^2}{2}\ge ab\) (đpcm)
c) ta có: \(\left(a+1\right)^2=a^2+2a+1\)
\(a\left(a+2\right)=a^2+2a\)
Vậy từ 2 điều trên => \(a\left(a+2\right)< \left(a+1\right)^2\)
d) \(m^2+n^2+2\ge2\left(m+n\right)\) (*)
<=>m2 - 2m +1 +n2 - 2n +1 \(\ge0\)
<=> \(\left(m-1\right)^2+\left(n-1\right)^2\ge0\) (1)
(1) đúng => (*) đúng
d) Bạn ấy giải rồi ,mình không giải nữa
e) Theo BĐT cauchy ta có: \(\dfrac{a^2+b^2}{2}\ge ab\Rightarrow\dfrac{a^2+b^2}{ab}\ge2\)
\(\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge2\Leftrightarrow\left(\dfrac{a}{b}+1\right)+\left(\dfrac{b}{a}+1\right)\ge4\)
\(\Leftrightarrow\dfrac{a+b}{b}+\dfrac{a+b}{a}\ge4\)
\(\Rightarrow\left(a+b\right)\left(\dfrac{1}{b}+\dfrac{1}{a}\right)\ge4\) (đpcm)
Vậy..........
e)
\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng)
=> ĐPCM
5. phân tích ra : \(1+\dfrac{a}{b}+\dfrac{b}{a}+1\)
áp dụng bđ cosy
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
=> đpcm
6. \(x^2-x+1=x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
hay với mọi x thuộc R đều là nghiệm của bpt
7.áp dụng bđt cosy
\(a^4+b^4+c^4+d^4\ge2\sqrt{a^2.b^2.c^2.d^2}=4abcd\left(đpcm\right)\)
Ta có: \(a^2+ab+b^2\)
\(=\left(a+b\right)^2-ab\ge\left(a+b\right)^2-\frac{\left(a+b\right)^2}{4}=\frac{3\left(a+b\right)^2}{4}\)
\(\Rightarrow\sqrt{a^2+ab+b^2}\ge\sqrt{\frac{3\left(a+b\right)^2}{4}}=\frac{\sqrt{3}}{2}\left(a+b\right)\)
Tương tự, ta có: \(\sqrt{b^2+bc+c^2}\ge\frac{\sqrt{3}}{2}\left(b+c\right)\)
\(\sqrt{c^2+ca+a^2}\ge\frac{\sqrt{3}}{2}\left(c+a\right)\)
Do đó ta có: \(Q\ge\frac{\sqrt{3}}{2}\left(a+b+b+c+c+a\right)=\sqrt{3}\) ( Do a+b+c=1)
Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)
Bài 2:
a) \(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(A=\dfrac{1}{abc}\left(a^3+b^3+c^3\right)\)
\(A=\dfrac{1}{abc}\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]\)
Vì \(a+b+c=0\)
Nên a + b = -c (1)
Thay (1) vào A, ta được:
\(A=\dfrac{1}{abc}\left[\left(-c\right)^3-3ab\left(-c\right)+c^3\right]\)
\(A=\dfrac{1}{abc}.3abc\)
\(A=3\)
b) \(B=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)
\(B=\dfrac{a^2}{a^2-\left(b^2+c^2\right)}+\dfrac{b^2}{b^2-\left(c^2+a^2\right)}+\dfrac{c^2}{c^2-\left(a^2+b^2\right)}\)
Vì \(a+b+c=0\)
Nên b + c = -a
=> ( b + c )2 = (-a)2
=> b2 + c2 + 2bc = a2
=> b2 + c2 = a2 - 2bc (1)
Tương tự ta có: c2 + a2 = b2 - 2ac (2)
a2 + b2 = c - 2ab (3)
Thay (1), (2) và (3) vào B, ta được:
\(B=\dfrac{a^2}{a^2-\left(a^2-2bc\right)}+\dfrac{b^2}{b^2-\left(b^2-2ac\right)}+\dfrac{c^2}{c^2-\left(c^2-2ab\right)}\)
\(B=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ac}+\dfrac{c^2}{c^2-c^2+2ab}\)
\(B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)
\(B=\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}\)
\(B=\dfrac{1}{2abc}\left(a^3+b^3+c^3\right)\)
Mà \(a^3+b^3+c^3=3abc\) ( câu a )
\(\Rightarrow B=\dfrac{1}{2abc}.3abc\)
\(\Rightarrow B=\dfrac{3}{2}\)
Bài 1:
a) GT: abc = 2
\(M=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)
\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{abc+2cb+2b}\)
\(M=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2+2cb+2b}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2\left(1+cb+b\right)}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(M=\dfrac{1+b+bc}{bc+b+1}\)
\(M=1\)
b) GT: abc = 1
\(N=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(N=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{cb}{b\left(ac+c+1\right)}\)
\(N=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{bc}{abc+bc+b}\)
\(N=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(N=\dfrac{1+b+bc}{bc+b+1}\)
\(N=1\)
Bài 1:
Áp dụng BĐt cauchy dạng phân thức:
\(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\ge\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3x+3y}=4\)
dấu = xảy ra khi 2x+y=x+2y <=> x=y
Bài 2:
ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{4^2}{a+b+c+d}=\dfrac{16}{a+b+c+d}\)(theo BĐt cauchy-schwarz)
\(\Rightarrow\dfrac{1}{a+b+c+d}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\)
Áp dụng BĐT trên vào bài toán ta có:
\(A=\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)\(A\le\dfrac{1}{16}.4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
......
dấu = xảy ra khi a=b=c
Bài 2:
Áp dụng BĐT cauchy cho 2 số dương:
\(a^2+1\ge2a\)
\(\Leftrightarrow\dfrac{a}{a^2+1}\le\dfrac{a}{2a}=\dfrac{1}{2}\)
thiết lập tương tự:\(\dfrac{b}{b^2+1}\le\dfrac{1}{2};\dfrac{c}{c^2+1}\le\dfrac{1}{2}\)
cả 2 vế các BĐT đều dương ,cộng vế với vế,ta có dpcm
dấu = xảy ra khi a=b=c=1