cho a+b=2. cmr a4+b4>=2.(dùng bđt bunhiacopxki)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2+7\right)\left(2a^2+\dfrac{7}{b^2}\right)\ge\left(2a+\dfrac{7}{b}\right)^2\)
\(\Rightarrow\sqrt{2a^2+\dfrac{7}{b^2}}\ge\dfrac{1}{3}\left(2a+\dfrac{7}{b}\right)\)
Tương tự: \(\sqrt{2b^2+\dfrac{7}{c^2}}\ge\dfrac{1}{3}\left(2a+\dfrac{7}{c}\right)\) ; \(\sqrt{2c^2+\dfrac{7}{a^2}}\ge\dfrac{1}{3}\left(2c+\dfrac{7}{a}\right)\)
Cộng vế:
\(VT\ge\dfrac{1}{3}\left(2a+2b+2c+\dfrac{7}{a}+\dfrac{7}{b}+\dfrac{7}{c}\right)=2+\dfrac{7}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(VT\ge2+\dfrac{7}{9}.\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (do \(a+b+c=3\))
\(VT\ge2+\dfrac{7}{9}.\left(\sqrt{a}.\sqrt{\dfrac{1}{a}}+\sqrt{b}.\sqrt{\dfrac{1}{b}}+\sqrt{c}.\sqrt{\dfrac{1}{c}}\right)^2=9\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. BĐT ban đầu
<=> \(\left(\frac{1}{3}-\frac{b}{a+3b}\right)+\left(\frac{1}{3}-\frac{c}{b+3c}\right)+\left(\frac{1}{3}-\frac{a}{c+3a}\right)\ge\frac{1}{4}\)
<=>\(\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}\ge\frac{3}{4}\)
<=> \(\frac{a^2}{a^2+3ab}+\frac{b^2}{b^2+3bc}+\frac{c^2}{c^2+3ac}\ge\frac{3}{4}\)
Áp dụng BĐT buniacoxki dang phân thức
=> BĐT cần CM
<=> \(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ac\right)}\ge\frac{3}{4}\)
<=> \(a^2+b^2+c^2\ge ab+bc+ac\)luôn đúng
=> BĐT được CM
2) \(a+b+c\le ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)\(\Leftrightarrow\)\(\left(a+b+c\right)^2-3\left(a+b+c\right)\ge0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a+b+c-3\right)\ge0\)\(\Leftrightarrow\)\(a+b+c\ge3\)
ko mất tính tổng quát giả sử \(a\ge b\ge c\)
Có: \(3\le a+b+c\le ab+bc+ca\le3a^2\)\(\Leftrightarrow\)\(3a^2\ge3\)\(\Leftrightarrow\)\(a\ge1\)
=> \(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\le\frac{3}{1+2a}\le1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
theo bài ta có:
a + b + c = 0
=> a = -(b + c)
=> a2 = [-(b + c)]2
=> a2 = b2 + 2bc + c2
=> a2 - b2 - c2 = 2bc
=> ( a2 - b2 - c2)2 = (2bc)2
=> a4 + b4 + c4 - 2a2c2 + 2b2c2 - 2a2c2 = 4b2c2
=> a4 + b4 + c4 = 2a2c2 + 2b2c2 + 2a2c2
=> 2(a4 + b4 + c4) = a4 + b4 + c4 + 2a2c2 + 2b2c2 + 2a2c2
=> 2(a4 + b4 + c4) = (a2 + b2 + c2)2
=> 2(a4 + b4 + c4) = 1
=> a4 + b4 + c4 = \(\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ai giúp em với ạ
Bài này thầy em bảo dùng BĐT Bunhiacopxki
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y=-5\cdot\dfrac{1-cos2x}{2}+12sin2x+7\)
\(=-\dfrac{5}{2}+\dfrac{5}{2}\cdot cos2x+12\cdot sin2x+7\)
\(=12\cdot sin2x+\dfrac{5}{2}\cdot cos2x+\dfrac{9}{2}\)
\(=\dfrac{\sqrt{601}}{2}\cdot\left(\dfrac{12\cdot sin2x}{\dfrac{\sqrt{601}}{2}}+cos2x\cdot\dfrac{5}{2}\cdot\dfrac{2}{\sqrt{601}}\right)+\dfrac{9}{2}\)
\(=\dfrac{\sqrt{601}}{2}\cdot\left(sin2x\cdot cosa+cos2x\cdot sina\right)+\dfrac{9}{2}\)
\(=\dfrac{\sqrt{601}}{2}\cdot sin\left(2x+a\right)+\dfrac{9}{2}\)
\(-1< =sin\left(2x+a\right)< =1\)
=>\(\dfrac{-\sqrt{601}}{2}< =\dfrac{\sqrt{601}}{2}\cdot sin\left(2x+a\right)< =\dfrac{\sqrt{601}}{2}\)
=>\(\dfrac{-\sqrt{601}+9}{2}< =y< =\dfrac{\sqrt{601}+9}{2}\)
\(y_{min}\) khi sin(2x+a)=-1
=>\(2x+a=-\dfrac{pi}{2}+k2pi\)
=>\(2x=-\dfrac{pi}{2}-a+k2pi\)
=>\(x=-\dfrac{pi}{4}-\dfrac{a}{2}+kpi\)
\(y_{max}\) khi sin(2x+a)=1
=>\(2x+a=\dfrac{pi}{2}+k2pi\)
=>\(x=\dfrac{pi}{4}-\dfrac{a}{2}+kpi\)
có bđt x² + y² ≥ (x+y)²/2 (*)
cm: (*) <=> 2x²+2y² ≥ x²+y²+2xy <=> x²+y²-2xy ≥ 0 <=> (x-y)² ≥ 0 bđt đúng
dấu "=" khi x = y
ad bđt (*) vào bài toán:
a^4 + b^4 ≥ (a²+b²)²/2 ≥ [(a+b)²/2]²/2 = [(2²)/2]²/2 = 2 (đpcm) ; dấu "=" khi a = b = 1
Áp dụng BĐT Bunhiacopxki:
\(\left(a^2+b^2\right)\left(1+1\right)\ge\left(a+b\right)^2\)
\(\Rightarrow\left(a^4+b^4\right)\left(1+1\right)\ge\left(a^2+b^2\right)^2\ge\frac{\left(a+b\right)^2}{2}=2\)
Dấu "=" <=> a=b=1