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1a) a2 + b2 + c2 + 2ab + 2bc + 2ca + a2 + b2 + c2
= ( a2 + 2ab +b2 ) + ( a2 + 2ac + c2 ) + ( b2 + 2bc + c2 )
= ( a + b )2 + ( a + c )2 + ( b + c )2
1b) 2.( ac - ab - bc + b2 ) + 2.( bc - ba - ac + a2 ) + 2.( ba - bc - ca + c2 )
= 2ac - 2ab - 2bc + 2b2 + 2bc - 2ab - 2ac +2a2 + 2ab - 2bc - 2ac + 2c2
= 2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc
= ( a2 - 2ab + b2 ) + (a2 - 2ac + c2 ) + (b2 - 2bc + c2 )
= (a-b)2 + (a-c)2 + (b-c)2
ta có \(Q=\frac{a^2+2a+1}{2a^2+\left(1-a\right)^2}+...\)
\(=\frac{a^2+2a+1}{3a^2-2a+1}+...=\frac{1}{3}+\frac{\frac{8}{3}a+\frac{2}{3}}{3a^2-2a+1}+...\)
\(=1+\frac{\frac{8}{3}a+\frac{2}{3}}{3a^2-2a+1}+\frac{\frac{8}{3}b+\frac{2}{3}}{3b^2-2b+1}+\frac{\frac{8}{3}c+\frac{2}{3}}{3c^2-2c+1}\)
mà \(3a^2-2a+1=3\left(a-\frac{1}{3}\right)^2+\frac{2}{3}\ge\frac{2}{3}\)
=>\(\frac{\frac{8}{3}a+\frac{2}{3}}{3a^2-2a+1}\le\frac{\frac{8}{3}a+\frac{2}{3}}{\frac{2}{3}}=\frac{3}{2}\left(\frac{8}{3}a+\frac{2}{3}\right)=4a+1\)
tương tự mấy cái kia rồi + vào, ta có
\(Q\le1+4\left(a+b+c\right)+3=8\)
dấu = xảy ra <=>a=b=c=1/3
^_^
a) Ta có: \(a^2-1\le0;b^2-1\le0;c^2-1\le0\)
\(\Rightarrow\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\le0\)
\(a^2+b^2+c^2\le1+a^2b^2+b^2c^2+c^2a^2-a^2b^2c^2\le1+a^2b^2+b^2c^2+c^2a^2\) ( vì \(abc\ge0\) )
Có \(b-1\le0\Rightarrow a^2b\sqrt{b}\left(b-1\right)\le0\Rightarrow a^2b^2\le a^2b\sqrt{b}\)
Tương tự: \(\hept{\begin{cases}b^2c^2\le b^2c\sqrt{c}\\c^2a^2\le c^2a\sqrt{a}\end{cases}\Rightarrow dpcm}\)
Ta có:
\(3\left(a^2+b^2+c^2\right)-3\left(a^2b+b^2c+c^2a\right)\)
= \(\left(a+b+c\right)\left(a^2+b^2+c^2\right)-3\left(a^2b+b^2c+c^2a\right)\)\(=a^3+ab^2+ac^2+a^2b+b^3+bc^2+ca^2+b^2c+c^3\)\(-3\left(a^2b+b^2c+c^2a\right)\)
\(=a^3+b^3+c^3+ab^2+bc^2+ca^2-2a^2b-2b^2c-2c^2a\)
\(=\left(a^3-2a^2b+ab^2\right)+\left(b^3-2b^2c+bc^2\right)+\left(c^3-2c^2a+ca^2\right)\)
\(=a\left(a-b\right)^2+b\left(b-c\right)^2+c\left(c-a\right)^2\)
Mà \(a,b,c>0\)
\(\Rightarrow a\left(a-b\right)^2+b\left(b-c\right)^2+c\left(c-a\right)^2\ge0\)
\(\Rightarrow\)\(3\left(a^2+b^2+c^2\right)\ge3\left(a^2b+b^2c+c^2a\right)\)
Lại có:
\(\left(a^2+b^2+c^2\right)^2+3\left(a^2+b^2+c^2\right)\ge6\left(a^2b+b^2c+c^2a\right)\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2\ge3\left(a^2b+b^2c+c^2a\right)\)<đpcm>
bài trên mk làm sai rồi, mong mọi người thông cảm và nghĩ cách khác nha
Không mất tính tổng quát ta giả sử \(a\ge b\ge c\)
Đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\a-c=z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}z\ge x\ge0\\z\ge y\ge0\end{matrix}\right.\)
Ta có:
\(x^2+y^2+z^2=\left(x-y\right)^2+\left(x+z\right)^2+\left(y+z\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2+2xz+2yz-2xy=0\)
\(\Leftrightarrow z^2+2xz+2yz+\left(x-y\right)^2=0\)
Vì \(\Rightarrow\left\{{}\begin{matrix}z\ge x\ge0\\z\ge y\ge0\end{matrix}\right.\)
\(\Rightarrow z^2+2xz+2yz+\left(x-y\right)^2\ge0\)
Dấu = xảy ra khi \(x=y=z=0\)
Hay \(a=b=c\)
\(VT=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2-4ab-4bc-4ca\)
\(VP=\left[\left(a+b\right)-2c\right]^2+\left[\left(b+c\right)-2a\right]^2+\left[\left(c+a\right)-2b\right]^2\)
\(=\left(a+b\right)^2-4\left(a+b\right)c+4c^2+\left(b+c\right)^2-4\left(b+c\right)a+4a^2+\left(a+c\right)^2-4\left(a+c\right)b+4b^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2-4\left(a+b\right)c+4c^2-4\left(b+c\right)a+4a^2-4\left(a+c\right)b+4b^2\)
Nhìn vào thấy 2 vế có \(\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\) rút gọn luôn thì được
\(-4ab-4bc-4ca=-4\left(a+b\right)c+4c^2-4\left(b+c\right)a+4a^2-4\left(a+c\right)b+4b^2\)
\(\Rightarrow ab-\left(a+b\right)c+c^2+bc-\left(b+c\right)a+a^2+ac-\left(a+c\right)c+b^2=0\)
\(\Rightarrow ab-ac-bc+c^2+bc-ab-ac+a^2+ac-ab-bc+b^2=0\)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Xảy ra khi \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Rightarrow a=b=c\)
4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)
=> (a+d)2 - (b+c)2= (a-d)2 - (c-b)2
=> a2+ d2+ 2ad - b2- c2- 2bc=a2 + d2 - 2ad - c2-b2+2bc
Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)
1) a2+b2+c2+3=2(a+b+c) =>(a-1)2+(b-1)2+(c-1)2=0
=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm