Chứng minh mà ko xảy ra dấu '' = '' a^2+b^4+c^2+1>2(ab+bc-ca)
K
Khách
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NH
2
H
6 tháng 4 2021
|a| + |b| >= |a+b|
<=> (|a|+|b|)^2 >= |a+b|^2
<=> a^2+b^2 +2|ab| >= a^2+b^2+2ab
<=> |ab| >= ab (luôn đúng)
Dấu = xảy ra khi a,b cùng dấu
KT
10 tháng 7 2018
a) Áp dụng BĐT AM-GM ta có:
\(x+y\ge2\sqrt{xy}\)
\(\Rightarrow\)\(\frac{x+y}{2}\ge\sqrt{xy}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y\)
b) Áp dụng BĐT AM-GM ta có:
\(\frac{\sqrt{x}}{\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{x}}\ge2\sqrt{\frac{\sqrt{x}}{\sqrt{y}}.\frac{\sqrt{y}}{\sqrt{x}}}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y\)
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. Dấu "=" xảy ra khi nào?
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 5 2021
Cách đơn giản nhất là sử dụng phép biến đổi tương đương:
BĐT đã cho tương đương:
\(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\)
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Vậy BĐT đã cho đúng.
Dấu "=" xảy ra khi x=y
NT
Nguyễn thành Đạt
CTVHS
20 tháng 3 2023
3.1
Xét hiệu :
\(\left(\dfrac{a+b}{2}\right)^2-ab=\dfrac{a^2+2ab+b^2}{4}-\dfrac{4ab}{4}\)
\(=\dfrac{a^2-2ab+b^2}{4}=\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\in R\)
Vậy \(\left(\dfrac{a+b}{2}\right)^2\ge ab,\forall a,b\in R\)
Dấu bằng xảy ra : \(\Leftrightarrow a=b\)
3.2
Áp dụng kết quả của câu 3.1 vào câu 3.2 ta được:
\(\left(a+b+c\right)^2=[a+\left(b+c\right)]^2\ge4a\left(b+c\right)\)
Mà : \(a+b+c=1\left(gt\right)\)
nên : \(1\ge4a\left(b+c\right)\)
\(\Leftrightarrow b+c\ge4a\left(b+c\right)^2\) ( vì a,b,c không âm nên b+c không âm )
Mà : \(\left(b+c\right)^2\ge4bc\Leftrightarrow\left(b-c\right)^2\ge0,\forall b,c\in N\)
\(\Rightarrow b+c\ge16abc\)
Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}a=b+c\\b=c\end{matrix}\right.\Leftrightarrow b=c=\dfrac{1}{4};a=\dfrac{1}{2}\)
KK
0
OT
1
29 tháng 6 2023
1<=x<=3
=>(x-1)>=0 và (x-3)<=0
=>(x-1)(x-3)<=0
=>x^2-4x+3<=0
=>x^2+3<=4x
Dấu = xảy ra khi x=1 hoặc x=3
NN
1 tháng 4 2017
c) Áp dụng BĐT cô si cho 2 hai số dương \(a;b\) ta có:
\(a+b\ge2\sqrt{ab}\)
\(\frac{1}{a}+\frac{1}{b}\ge\frac{1}{\sqrt{ab}}\)
\(\Rightarrow\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
Dấu "=" xảy ra khi \(\Leftrightarrow a=b\)
LN
0
23 tháng 5 2021
Áp dụng cosi có:
\(\sqrt{x\left(2x+y\right)}=\dfrac{1}{\sqrt{3}}\sqrt{3x\left(2x+y\right)}\le\dfrac{1}{\sqrt{3}}.\dfrac{5x+y}{2}\)
\(\sqrt{y\left(2y+x\right)}\le\dfrac{1}{\sqrt{3}}.\dfrac{5y+x}{2}\)
\(\Rightarrow P\ge\dfrac{x+y}{\dfrac{1}{2\sqrt{3}}\left(6x+6y\right)}=\dfrac{\sqrt{3}}{3}\)
Dấu = xảy ra khi x=y
23 tháng 5 2021
Bài này áp dụng bunhia :v
Áp dụng bunhia với 2 cặp số `(sqrtx,sqrty),(sqrt{2x+y},sqrt{2y+x})`
`(x+y)(2x+y+2y+x)>=(sqrt{x(2x+y)}+sqrt{y(2y+x)})^{2}`
`<=>3(x+y)^{2}>=(sqrt{x(2x+y)}+sqrt{y(2y+x)})^{2}`
`=>sqrt{x(2x+y)}+sqrt{(2y+x)}<=sqrt3(x+y)`
`=>P>=1/sqrt3`
Dấu "="`<=>x=y`