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\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3\left(ab+bc+ca\right)\Leftrightarrow a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2\left(a^2+b^2+c^2\right)=2ab+2bc+2ca\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0mà:\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Rightarrow a=b=c\)
a, Ta xét hiệu \(\frac{a^2+b^2}{2}-\left(\frac{a+b}{2}\right)^2\)
\(=\frac{2\left(a^2+b^2\right)}{4}-\frac{a^2+2ab+b^2}{4}=\frac{1}{4}\left(2a^2+2b^2-a^2-b^2-2ab\right)=\frac{1}{4}\left(a+b\right)^2\ge0\)
Vậy \(\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2.\)
Dấu "="xảy ra khi a = b.
b, Ta xét hiệu:
\(\frac{a^2+b^2+c^2}{3}-\left(\frac{a+b+c}{3}\right)^2=\frac{1}{9}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)
Vậy \(\frac{a^2+b^2+c^2}{3}\ge\left(\frac{a+b+c}{3}\right)^2\)
Dấu "=" xảy ra khi a = b = c.
Cách khác:\(a^2+b^2\ge2ab\Leftrightarrow a^2+b^2+a^2+b^2\ge\left(a+b\right)^2\Leftrightarrow a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\)
\(\frac{a^2+b^2}{2}\ge\frac{\frac{\left(a+b\right)^2}{2}}{2}=\left(\frac{a+b}{2}\right)^2\)
Dấu " = " xảy ra <=> a=b
\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=ab\left(a-b\right)+bc\left[\left(b-a\right)-\left(c-a\right)\right]+ca\left(c-a\right)\)
\(=ab\left(a-b\right)-bc\left(a-b\right)-bc\left(c-a\right)+ca\left(c-a\right)\)
\(=\left(a-b\right)\left(ab-bc\right)-\left(c-a\right)\left(bc-ca\right)\)
\(=b\left(a-b\right)\left(a-c\right)-c\left(c-a\right)\left(b-a\right)\)
\(=b\left(a-b\right)\left(a-c\right)-c\left(a-c\right)\left(a-b\right)\)
\(=\left(a-c\right)\left(a-b\right)\left(b-c\right)\)
Câu 1:
a) a(a+2b)3 - b(2a+b)3 = a( a3 + 6a2b + 12ab2 + 8b2) - b
= a( a3 + 6a2b + 12ab2 + 8b3) - b( 8a3 + 12a2b + 6ab2 + b3)
= a4 + 6a3b + 12a2b2 + 8ab3 - 8a3b -12a2b2 - 6ab3 - b4
= a4 - 2a3b + 2ab3 - b4
= (a - b )(a + b)(a2 +b2) - 2ab(a - b)(a + b)
= (a - b )(a + b)(a2 +b2 -2ab)
= (a - b )3(a + b)
a) \(a^2+b^2+c^2\ge ab+bc+ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)( luôn đúng )
Dấu "=" \(\Leftrightarrow a=b=c\)
b) \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
+) vế 1 bđt \(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac\)( CMTT câu a )
+) vế 2 bđt \(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ac\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac\)( CMTT câu a )
Từ đây ta có đpcm
Dấu "=" \(\Leftrightarrow a=b=c\)
c) \(a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\)
\(\Leftrightarrow a^2-ab+b^2\ge ab\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)( luôn đúng )
Dấu "=" \(\Leftrightarrow a=b\)