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1)Áp dụng Bđt Am-Gm \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}=2\)
2)Áp dụng Am-Gm \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab;b^2+c^2\ge2bc;a^2+c^2\ge2ca\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
=>ĐPcm
3)(a+b+c)2\(\ge\)3(ab+bc+ca)
=>a2+b2+c2+2ab+2bc+2ca\(\ge\)3ab+3bc+3ca
=>a2+b2+c2-ab-bc-ca\(\ge\)0
=>2a2+2b2+2c2-2ab-2bc-2ca\(\ge\)0
=>(a2-2ab+b2)+(b2-2bc+c2)+(c2-2ac+a2)\(\ge\)0
=>(a-b)2+(b-c)2+(c-a)2\(\ge\)0
4)đề đúng \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)
a: \(=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{a^2+b^2+c^2-ab-ac-bc}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)}{a^2+b^2+c^2-ab-ac-bc}\)
=a+b+c
e: \(=\dfrac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{a\left(b^2-c^2\right)-b\left(b^2-c^2\right)}\)
\(=\dfrac{ab\left(a-b\right)+c\left(b-a\right)\left(b+a\right)+c^2\left(a-b\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{\left(a-b\right)\left(ab-ac-bc+c^2\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{a\left(b-c\right)-c\left(b-c\right)}{\left(b-c\right)\left(b+c\right)}=\dfrac{a-c}{b+c}\)
Câu a/ Thì chứng minh ở dưới rồi nhé e
b/ Ta cần chứng minh
\(2\left(a^2b^2+b^2c^2+c^2a^2\right)=2\left(ab+bc+ca\right)^2\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\)
\(\Leftrightarrow2abc\left(a+b+c\right)=0\)(đúng)
=> ĐPCM
c/ Ta có
\(\frac{\left(a^2+b^2+c^2\right)^2}{2}=\frac{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)}{2}=a^4+b^4+c^4\)
Cái này là áp dụng câu a vô nhé e
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
Ta có:
a)
\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2a^2b^2-2a^2c^2-2b^2c^2\)
\(=\left[\left(a+b+c\right)^2-2ab-2ac-2bc\right]^2-2a^2b^2-2b^2c^2-2a^2c^2\)
\(=4\left[ab+ac+bc\right]^2-2a^2b^2-2b^2c^2-2a^2c^2\)
\(=4\left(ab\right)^2+4\left(ac\right)^2+4\left(bc\right)^2-8abc\left(a+b+c\right)-2a^2b^2-2b^2c^2-2a^2c^2\)
\(=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
b)\(=2\left(ab+bc+ac\right)^2-4\left(abbc+abca+bcca\right)\)
\(=2\left(ab+bc+ac\right)^2-4abc\left(a+b+c\right)=2\left(ab+bc+ac\right)^2\)
c) \(\frac{\left(a^2+b^2+c^2\right)^2}{2}=\frac{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)}{2}=\frac{a^4+b^4+c^4+a^4+b^4+c^4}{2}\)
\(=a^4+b^4+c^4\)
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