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Ta chứng minh BĐT sau với các số dương:
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Thật vậy, BĐT 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)
Áp dụng:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ; \(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) ; \(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)
Cộng vế với vế:
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
b.
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\Rightarrow\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)
\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)
Cộng vế với vế (1); (2) và (3):
\(\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
1/ \(a^2-b^2+c^2\ge\left(a-b+c\right)^2\)
\(\Leftrightarrow bc-ac-b^2+ab\ge0\)
\(\Leftrightarrow\left(bc-ac\right)+\left(ab-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\ge0\)(đúng)
Vì \(\hept{\begin{cases}a\ge b\\b\ge c\end{cases}}\)
2/ \(a^2-b^2+c^2-d^2\ge\left(a-b+c-d\right)^2\)
\(\Leftrightarrow-d^2+cd-bd+ad+bc-ac-b^2+ab\ge0\)
\(\Leftrightarrow\left(dc-d^2\right)+\left(ad-bd\right)+\left(bc-ac\right)+\left(ba-b^2\right)\ge0\)
\(\Leftrightarrow d\left(c-d\right)+d\left(a-b\right)+\left(a-b\right)\left(b-c\right)\ge0\)
Đúng vì \(a\ge b\ge c\ge d\ge0\)
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
Định đi ngủ mà chợt nhớ lúc chiều có hứa là làm giúp chủ tus nên h phải làm =)))
c) Áp dụng BĐT Cauchy-schwars ta có:
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{\left(a+b+b\right)^2}{a+b+c}=a+b+c\)
đpcm
a) \(2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)
<=> \(a^4+b^4\ge ab\left(a^2+b^2\right)\)
Ta có: \(a^4+b^4\ge\frac{\left(a^2+b^2\right)^2}{2}=\frac{a^2+b^2}{2}.\left(a^2+b^2\right)\ge ab\left(a^2+b^2\right)\) với mọi a, b
Vậy \(2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)
Dấu "=" xảy ra <=> a = b
b) \(3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)(1)
<=> \(2\left(a^4+b^4+c^4\right)\ge ab^3+ac^3+ba^3+bc^3+ca^3+cb^3\)
<=> \(\left(a^4+b^4\right)+\left(b^4+c^4\right)+\left(c^4+a^4\right)\ge ab\left(a^2+b^2\right)+bc\left(b^2+c^2\right)+ac\left(a^2+c^2\right)\) đúng áp dụng câu a
Vậy (1) đúng
Dấu "=" xảy ra <=> a = b = c.
1a)\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
b)\(\dfrac{a^2+b^2+c^2}{3}\ge\dfrac{\left(a+b+c\right)^2}{9}\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)
2a)\(a^2+\dfrac{b^2}{4}\ge ab\)
\(\Leftrightarrow a^2-ab+\dfrac{b^2}{4}\ge0\)
\(\Leftrightarrow a^2-2\cdot\dfrac{1}{2}b\cdot a+\left(\dfrac{1}{2}b\right)^2\ge0\)
\(\Leftrightarrow\left(a-\dfrac{1}{2}b\right)^2\ge0\)(luôn đúng)
b)Đã cm
c)\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng)
Dấu bằng xảy ra khi a=b=1
a/CM: \(\left(\frac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\frac{a+b}{2}\ge\sqrt{ab}\)
\(\Leftrightarrow a+b\ge2\sqrt{ab}\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) ( luôn đúng với mọi a,b>0)
CM: \(\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2\)
\(\Leftrightarrow\frac{2\left(a^2+b^2\right)}{4}\ge\frac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow a^2+b^2\ge2ab\) ( luôn đúng)
b/CM: \(\frac{a^3+b^3}{2}\ge\left(\frac{a+b}{2}\right)^3\)
\(\Leftrightarrow\frac{4\left(a^3+b^3\right)}{8}\ge\frac{\left(a+b\right)^3}{8}\)
\(\Leftrightarrow3\left(a^3+b^3\right)\ge3a^2b+3ab^2\)
\(\Leftrightarrow a^2\left(a-b\right)+b^2\left(b-a\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) ( luôn đúng với mọi a,b>0)
c/CM: \(a^4+b^4\ge a^3b+ab^3\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+b^2+ab\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+\frac{2ab}{2}+\frac{b^2}{4}+\frac{3b^2}{4}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\right)\ge0\) ( luôn đúng)
d/Ta xét hiệu: \(a^4-4a+3\)
\(=a^4-2a^2+1+2a^2-4a+2\)
\(=\left(a-1\right)^2+2\left(a-1\right)^2\ge0\)
Suy ra BĐT luôn đúng
e/Ta xét hiệu:( Làm nhanh)
\(a^3+b^3+c^3-3abc\)\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right)\ge0\)
f/Ta có: \(\frac{a^6}{b^2}-a^4+\frac{a^2b^2}{4}+\frac{b^6}{a^2}-b^4+\frac{a^2b^2}{4}\)
\(=\left(\frac{a^3}{b}-\frac{ab}{2}\right)^2+\left(\frac{b^3}{a}-\frac{ab}{2}\right)^2\ge0\)(1)
Mà \(\frac{a^2b^2}{4}+\frac{a^2b^2}{4}\ge0\)(2)
Lấy (1) trừ (2) được: \(\frac{a^6}{b^2}+\frac{b^6}{a^2}-a^4-b^4\ge0\RightarrowĐPCM\)
g/Làm rồi..xem lại trong trang cá nhân
h/Xét hiệu có: \(\left(a^5+b^5\right)\left(a+b\right)-\left(a^4+b^4\right)\left(a^2+b^2\right)\)
\(=a^5b+ab^5-a^2b^4-a^4b^2\)
\(=a^4b\left(a-b\right)-ab^4\left(a-b\right)\)
\(=ab\left(a^2-b^2\right)\left(a-b\right)\)
\(=ab\left(a+b\right)\left(a-b\right)^2\ge0\forall ab>0\)
Suy ra ĐPCM
a/ a3 - b3 \(\ge\)3a2b - 3ab2
<=> a3 - b3 - 3ab(a - b) \(\ge0\)
<=> (a - b)3 \(\ge0\)(đúng)
b/ \(a^2+b^2+c^2\ge a+b+c-\frac{3}{4}\)
\(\Leftrightarrow\left(a^2-a+\frac{1}{4}\right)+\left(b^2-b+\frac{1}{4}\right)+\left(c^2-c+\frac{1}{4}\right)\ge0\)
\(\Leftrightarrow\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\left(c-\frac{1}{2}\right)^2\ge0\)
=> ĐPCM
lần sau gõ từ với ko có mất thời gian bn ký hiệu \(\gamma\) ng` ta hiểu thành kí hiệu tia Gamma thì sao