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(a+b+c+d)2\(\ge\frac{8}{3}\)(ab+ac+ad+bc+bd+cd)
<=>(a+b)2+2(a+b)(c+d)+(c+d)2\(\ge\).....
<=>a2+b2+c2+d2+2(ab+ac+ad+bc+bd+cd)\(\ge\)....
<=>3a2+3b2+3c2+3d2+6(ab+ac+ad+bc+bd+cd)\(\ge\)8(ab+ac+ad+bc+bd+cd)
<=> 3a2+3b2+3c2+3d2-2ab -2ac-2bc-2ad-2bd-2cd\(\ge\)0
<=> (a2-2ab+b2)+(a2-ac+c2)+(a2-2ad+d2)+(b2-2bc+c2)+(b2-2bd+d2)+(c2-2cd+d2)>=0
<=> (a-b)2+(a-c)2+(a-d)2+(b-c)2+(b-d)2+(c-d)2>=0 (DPCM)
Dau ''='' xay ra khi a=b=c=d
A) Ta có :
Vế phải = ( a + b ) ( a2 - 2ab + b2 +ab )
= ( a + b ) ( a2 - ab + b2 )
= a3 + b3 = Vế trái ( điều phải chứng minh )
Chúc bạn học tốt ^^
Câu a) thôi nhé
Ta có (a+b) [(a-b)2+ab] = (a+b)(a2-ab-b2) = a3-a2b + ab2 + ba2 - ab2 +b3
Thu gọn lại ta được a3 + b3
1.
\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)
\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
Dấu "=" khi \(a=b=c\)
2.
\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" khi \(a=b=c=d\)
Ta có :
\(3\left(a^2+b^2+c^2+d^2\right)-2\left(ab+ac+ad+bc+bd+cd\right)\)
\(=\left(a-b\right)^2+\left(a-c\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(c-d\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2+d^2\ge\frac{2}{3}\left(ab+ac+ad+bc+bd+cd\right)\)
\(\Rightarrow\left(a+b+c+d\right)^2=a^2+b^2+c^2+d^2+2\left(ab+ac+ad+bc+bd+cd\right)\)
\(\ge\frac{8}{3}\left(ab+ac+ad+bc+bd+cd\right)\left(đpcm\right)\)
\(\left(a+b+c+d\right)^2\ge\frac{8}{3}\left(ab+ac+ad+bc+bd+cd\right)\)
\(\Leftrightarrow a^2+b^2+c^2+d^2+2\left(ab+ac+ad+bc+bd+cd\right)\ge\frac{8}{3}\left(ab+ac+ad+bc+bd+cd\right)\)
\(\Leftrightarrow3\left(a^2+b^2+c^2+d^2\right)+6\left(ab+ac+ad+bc+bd+cd\right)\ge8\left(ab+ac+ad+bc+bd+cd\right)\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(a^2-2ad+d^2\right)+\left(b^2-2bc+c^2\right)+\left(b^2-2bd+d^2\right)\)\(+\left(c^2-2cd+d^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(c-d\right)^2\ge0\) ( đúng )
=> Đpcm