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Câu 1:
Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2^2}-ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\) (1)
Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\) (2)
Từ (1) và (2) \(\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
5 , a3+b3+c3\(\ge\) 3abc
\(\Leftrightarrow\) a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc\(\ge\) 0
\(\Leftrightarrow\) (a+b)3+c3-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+b2+c2-ab-bc-ca)\(\ge0\) (1)
ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)
(a-b)2+(b-c)2+(c-a)2\(\ge0\)
<=> 2a2+2b2+2c2-2ac-2cb-2ab\(\ge0\)
<=>a2+b2+c2-ab-bc-ac\(\ge\) 0 (3)
Từ (1)(2)(3)=> pt luôn đúng
a)\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)
b,c tương tự
d)Áp dụng bđt AM-GM ta được
\(a^4+a^4+b^4+c^4\ge4\sqrt[4]{a^4a^4b^4c^4}=4a^2bc\)
TT\(\Rightarrow a^4+b^4+b^4+c^4\ge4ab^2c\)
\(a^4+b^4+c^4+c^4\ge4abc^2\)
Cộng vế theo vế ta được \(4\left(a^4+b^4+c^4\right)\ge4\left(a^2bc+ab^2c+abc^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4\ge abc\left(a+b+c\right)\left(đpcm\right)\)
d)
\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
\(\Leftrightarrow a^4+b^4+c^4-a^2bc-ab^2c-abc^2\ge0\)
\(\Leftrightarrow2a^4+2b^4+2c^4-2a^2bc-2ab^2c-2abc^2\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+2a^2b^2+\left(b^2-c^2\right)^2+2b^2c^2+\left(c^2-a^2\right)^2+2a^2c^2-2a^2bc-2b^2ac-2c^2ab\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2+\left(a^2b^2+b^2c^2-2b^2ac\right)+\left(b^2c^2+c^2a^2-2c^2abc\right)+\left(a^2b^2+c^2a^2-2a^2ab\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2+\left(ab-bc\right)^2+\left(bc-ac\right)^2+\left(ab-ac\right)^2\ge0\)
Luôn đúng với mọi a , b , c
Lời giải:
Sử dụng pp biến đổi tương đương:
a) \(\frac{a^2+b^2}{2}\geq \left(\frac{a+b}{2}\right)^2\)
\(\Leftrightarrow \frac{a^2+b^2}{2}\geq \frac{(a+b)^2}{4}\)
\(\Leftrightarrow 4(a^2+b^2)\geq 2(a+b)^2\Leftrightarrow 4(a^2+b^2)\geq 2(a^2+2ab+b^2)\)
\(\Leftrightarrow 2(a^2+b^2)\geq 4ab\Leftrightarrow 2(a^2+b^2-2ab)\geq 0\)
\(\Leftrightarrow 2(a-b)^2\geq 0\) (luôn đúng)
Do đó ta có đpcm. Dấu bằng xẩy ra khi $a=b$
c)
\(\frac{a^2+b^2+c^2}{3}\geq \left(\frac{a+b+c}{3}\right)^2\) \(\Leftrightarrow \frac{a^2+b^2+c^2}{3}\geq \frac{(a+b+c)^2}{9}\)
\(\Leftrightarrow 3(a^2+b^2+c^2)\geq (a+b+c)^2\)
\(\Leftrightarrow 3(a^2+b^2+c^2)\geq a^2+b^2+c^2+2(ab+bc+ac)\)
\(\Leftrightarrow 2(a^2+b^2+c^2)\geq 2(ab+bc+ac)\)
\(\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)\geq 0\)
\(\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2\geq 0\) (luôn đúng)
Do đó ta có đpcm. Dấu bằng xảy ra khi $a=b=c$
b) \(\frac{a^4+b^4}{2}\geq \left(\frac{a+b}{2}\right)^4\)
Áp dụng 2 lần BĐT phần a: \(\frac{a^4+b^4}{2}\geq \left(\frac{a^2+b^2}{2}\right)^2(1)\)
Và: \(\frac{a^2+b^2}{2}\geq \left(\frac{a+b}{2}\right)^2\Rightarrow \left(\frac{a^2+b^2}{2}\right)^2\geq \left(\frac{a+b}{2}\right)^4(2)\)
Từ \((1); (2)\Rightarrow \frac{a^4+b^4}{2}\geq \left(\frac{a+b}{2}\right)^4\) (đpcm)
Dấu bằng xảy ra khi \(a=b\)
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.
1.b
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-d\right)^2+\left(d-a\right)^2\ge0\) tong 4 so khong am luon dung
2 . ta có
\(\left(x-y\right)^2\ge0\)
<=> x2-2xy+y2 ≥ 0
<=> x2+4xy-2xy+y2 ≥ 4xy
<=> x2+2xy+y2 ≥ 4xy
<=> (x+y)2 ≥ 4xy
CMTT
(y+z)2 ≥ 4yz
(z+x)2 ≥ 4zx
nhân các vế của bđt ta có
[(x+y)(y+z)(z+x)]2 ≥ 64x2y2z2
<=> (x+y)(y+z)(z+x) ≥ 8xyz
1, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)
Từ (1), (2) và (3) suy ra:
\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)
<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\) \(\xrightarrow[]{}\) đpcm
5. a, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)
\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)
\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)
Từ (1),(2) và (3) suy ra:
\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)
<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)
mà x+y+z=3
=>\(x^2+y^2+z^2+3\ge2.3=6\)
<=> \(x^2+y^2+z^2\ge6-3=3\)
<=> \(A\ge3\)
Dấu "=" xảy ra khi x=y=z=1
Vậy GTNN của A=x2+y2+z2 là 3 khi x=y=z=1
b, Ta có: x+y+z=3
=> \(\left(x+y+z\right)^2=9\)
<=> \(x^2+y^2+z^2+2xy+2yz+2xz=9\)
<=> \(x^2+y^2+z^2=9-2xy-2yz-2xz\)
mà \(x^2+y^2+z^2\ge3\) (theo a)
=> \(9-2xy-2yz-2xz\ge3\)
<=> \(-2\left(xy+yz+xz\right)\ge3-9=-6\)
<=> \(xy+yz+xz\le\dfrac{-6}{-2}=3\)
<=> \(B\le3\)
Dấu "=" xảy ra khi x=y=z=1
Vậy GTLN của B=xy+yz+xz là 3 khi x=y=z=1
a.
\(2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)
\(\Leftrightarrow2a^4+2b^4\ge a^4+ab^3+a^3b+b^4\)
\(\Leftrightarrow a^4+b^4\ge ab^3+a^3b\)
\(\Leftrightarrow a^4-a^3b+b^4-ab^3\ge0\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^3-b^3\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)(*)
Mà \(a^2+ab+b^2=\left(a^2+2\cdot a\cdot\dfrac{1}{2}b+\dfrac{b^2}{4}\right)+\dfrac{3b^2}{4}=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\ge0\)
Suy ra (*) đúng => đpcm
Dấu "=" xảy ra khi a = b
b.
\(3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)
\(\Leftrightarrow3a^4+3b^4+3c^4\ge a^4+ab^3+ac^3+a^3b+b^4+bc^3+a^3c+b^3c+c^4\)
\(\Leftrightarrow2a^4+2b^4+2c^4\ge ab^3+a^3b+b^3c+bc^3+ca^3+c^3a\)
\(\Leftrightarrow\left(a^4+b^4\right)+\left(b^4+c^4\right)+\left(c^4+a^4\right)\ge\left(a^3b+ab^3\right)+\left(b^3c+bc^3\right)+\left(c^3a+ca^3\right)\)
Theo câu a. thì điều này đúng
Dấu "=" khi a=b=c