CMR: a=b=c nếu có một trong các điều kiện sau
a) a2+b2+c2 = ab+bc+ca
b) (a+b+c)2 = 3(a2+b2+c2)
c) (a+b+c)2 = 3(ab+bc+ca)
P/s: Cần gấp....~~
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`1)(a+b+c)^2=3(a^2+b^2+c^2)`
`<=>a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+3c^2`
`<=>2ab+2bc+2ca=2a^2+2b^2+2c^2`
`<=>a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0`
`<=>(a-b)^2+(b-c)^2+(c-a)^2=0`
Mà `(a-b)^2+(b-c)^2+(c-a)^2>=0`
Vậy dấu "=" xảy ra chỉ có thể là `a=b=c`
`2)(a+b+c)^2=3(ab+bc+ca)`
`<=>a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca`
`<=>a^2+b^2+c^2=ab+bc+ca`
`<=>2ab+2bc+2ca=2a^2+2b^2+2c^2`
`<=>a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0`
`<=>(a-b)^2+(b-c)^2+(c-a)^2=0`
Mà `(a-b)^2+(b-c)^2+(c-a)^2>=0`
Vậy dấu "=" xảy ra chỉ có thể là `a=b=c`
Vậy nếu `a=b=c` thì ....
c: Ta có: \(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
\(=a^4-2a^3b+2ab^3-b^4\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a-b\right)^3\cdot\left(a+b\right)\)
Sửa đề: 1+a^2;1+b^2;1+c^2
\(\dfrac{a}{\sqrt{1+a^2}}=\dfrac{a}{\sqrt{a^2+ab+c+ac}}=\sqrt{\dfrac{a}{a+b}\cdot\dfrac{a}{a+c}}< =\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
\(\dfrac{b}{\sqrt{1+b^2}}< =\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{b}{b+a}\right)\)
\(\dfrac{c}{\sqrt{1+c^2}}< =\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{a+b}\right)\)
=>\(A< =\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{3}{2}\)
Đặt \(P=\dfrac{a^3}{a^2+b^2+ab}+\dfrac{b^3}{b^2+c^2+bc}+\dfrac{c^3}{c^2+a^2+ca}\)
Ta có: \(\dfrac{a^3}{a^2+b^2+ab}=a-\dfrac{ab\left(a+b\right)}{a^2+b^2+ab}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^3b^3}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)
Tương tự: \(\dfrac{b^3}{b^2+c^2+bc}\ge\dfrac{2b-c}{3}\) ; \(\dfrac{c^3}{c^2+a^2+ca}\ge\dfrac{2c-a}{3}\)
Cộng vế:
\(P\ge\dfrac{a+b+c}{3}=673\)
Dấu "=" xảy ra khi \(a=b=c=673\)
a.
Theo BĐT tam giác: \(c< a+b\Rightarrow c^2< ac+bc\)
\(b< a+c\Rightarrow b^2< ab+bc\) ; \(a< b+c\Rightarrow a^2< ab+ac\)
Cộng vế với vế: \(a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
b.
Do a;b;c là 3 cạnh của tam giác nên: \(\left\{{}\begin{matrix}a+b-c>0\\b+c-a>0\\c+a-b>0\end{matrix}\right.\)
\(\left(a+b-c\right)\left(b+c-a\right)\le\dfrac{1}{4}\left(a+b-c+b+c-a\right)^2=b^2\)
Tương tự: \(\left(b+c-a\right)\left(a+c-b\right)\le c^2\) ; \(\left(a+b-c\right)\left(a+c-b\right)\le a^2\)
Nhân vế với vế:
\(\left(abc\right)^2\ge\left[\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\right]^2\)
\(\Leftrightarrow abc\ge\left(a+b-c\right)\left(c+a-b\right)\left(b+c-a\right)\)
a) a2 + b2 + c2 = ab + bc + ac
\(\Rightarrow\) a2 + b2 + c2 - ab - bc - ac = 0
\(\Rightarrow\) 2(a2 + b2 + c2 - ab - bc - ac) = 0
\(\Rightarrow\) a2 + a2 + b2 + b2 + c2 + c2 - 2ab - 2bc - 2ac = 0
\(\Rightarrow\) (a2 - 2ab + b2) + (a2 - 2ac + c2) + (b2 - 2bc + c2) = 0
\(\Rightarrow\) (a - b)2 + (a - c)2 + (b - c)2 = 0
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0\\a=c\\b=c\end{matrix}\right.\)
\(\Rightarrow\) a = b = c
b) (a + b + c)2 = 3(a2 + b2 + c2)
a2 + b2 + c2 + 2ab + 2bc + 2ac = 3a2 + 3b2 + 3c2
\(\Rightarrow\) 2ab + 2ac + 2bc = 2a2 + 2b2 + 2c2
\(\Rightarrow\) 0 = a2 + a2 + b2 + b2 + c2 + c2 - 2ab - 2bc - 2ac
Hay: a2 + a2 + b2 + b2 + c2 + c2 - 2ab - 2bc - 2ac = 0
\(\Rightarrow\)(a2 - 2ab + b2) + (a2 - 2ac + c2) + (b2 - 2bc + c2) = 0
\(\Rightarrow\) (a - b)2 + (a - c)2 + (b - c)2 = 0
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0\\a=c\\b=c\end{matrix}\right.\)
\(\Rightarrow\) a = b = c
c) (a + b + c)2 = 3(ab + bc + ac)
a2 + b2 + c2 + 2ab + 2ac + 2bc = 3ab + 3bc + 3ac
\(\Rightarrow\) a2 + b2 + c2 = ab + ac + bc2
\(\Rightarrow\) 2(a2 + b2 + c2) = 2(ab + ac + bc)
\(\Rightarrow\) 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ac
\(\Rightarrow\) a2 + a2 + b2 + b2 + c2 + c2 - 2ab - 2bc - 2ac = 0
\(\Rightarrow\) (a2 - 2ab + b2) + (a2 - 2ac + c2) + (b2 - 2bc + c2) = 0
\(\Rightarrow\) (a - b)2 + (a - c)2 + (b - c)2 = 0
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0\\a=c\\b=c\end{matrix}\right.\)
\(\Rightarrow\) a = b = c
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a)Cmr : Nếu : a2 + b2 + c2 = ab + bc + ac thì a = b =c
Bài làm
2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ac = 0
=> ( a2 - 2ab + b2) + ( a2 - 2ac + c2) + ( b2 - 2bc + c2) =0
= > ( a - b)2 + ( a - c)2 + ( b -c)2 = 0
Vậy :
* ( a - b)2 = 0
* ( a - c)2 =0
* (b -c)2 =0
Suy ra :
* a =b
* a =c
* b = c
Suy ra : a = b =c ( đpcm)