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a) \(a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)
=> ĐPCM
b) \(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=3a^2+3b^2+3c^2\)
\(\Leftrightarrow-2a^2-2b^2-2c^2+2ab+2bc+2ac=0\)
\(\Leftrightarrow-\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0\)
\(\Leftrightarrow-\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)
=> ĐPCM
c) \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=3ab+3bc+3ac\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)
=> ĐPCM
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(=a^2+b^2+c^2+2\left(ab+bc+ac\right)=3\left(a^2+b^2+c^2\right)\)
\(\Rightarrow2\left(ab+bc+ac\right)=3\left(a^2+b^2+c^2\right)-\left(a^2+b^2+c^2\right)\)
\(\Rightarrow2\left(ab+bc+ac\right)=2\left(a^2+b^2+c^2\right)\)
\(\Rightarrow ab+bc+ac=a^2+b^2+c^2\)
\(\Rightarrow a=b=c\left(đpcm\right)\)
@Ngọc Minh Dương
Cách tách ra là cách của người học toán mức TB
Đề bắt C/m nhé
VT=0 hiển nhiên
VP=\(3\left[\left(a^2-ab\right)+\left(b^2-bc\right)+\left(c^2-ca\right)\right]=3\left[a\left(a-b\right)+b\left(b-c\right)+c\left(c-a\right)\right]=3.\left[0+0+0\right]=3.0=0\)VT=VP=0
Lưu Hiền cái cách của bạn --> đúng cái đề này không cần hỏi >>> cái người hỏi cần cách làm bằng bộ não không phải làm = chân tay
Có :
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(=a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ac\)
\(=2a^2+2b^2+2c^2-2ab-2bc-2ab\)
\(3\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=3a^2+3b^2+3c^2-3ab-3bc-3ac\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ab=3a^2+3b^2+3c^2-3ab-3bc-3ac\)
Trừ cả 2 vế đi \(2a^2+2b^2+2c^2-2ab-2ac-2bc;\)có :
\(\Rightarrow a^2+b^2+c^2-bc-ca-ac=0\)
\(\Rightarrow2\left(a^2+b^2+c^2-bc-ca-ac\right)=0.2\)
\(\Rightarrow\left(a^2+b^2-2ab\right)+\left(b^2+c^2-2bc\right)+\left(a^2+c^2-2ab\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Mà \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}}\)
\(\Rightarrow a-b=b-c=c-a=0\)
\(\Rightarrow a=b=c\)
Vậy ...
a,
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=2\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\)
b,
\(a+b+c=2p\Leftrightarrow p=\dfrac{a+b+c}{2}\)
\(\Leftrightarrow\left(p-a\right)^2+\left(p-b\right)^2+\left(p-c\right)^2=3p^2-2pa-2pb-2pc+a^2+b^2+c^2\)
\(=3\left(\dfrac{a+b+c}{2}\right)^2-2\cdot\dfrac{a+b+c}{2}\cdot a-2\cdot\dfrac{a+b+c}{2}\cdot b-2\cdot\dfrac{a+b+c}{2}\cdot c+a^2+b^2+c^2\)
\(=3p^2-\left(a+b+c\right)^2+a^2+b^2+c^2=3p^2-4p^2+a^2+b^2+c^2=a^2+b^2+c^2-p^2\)
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 \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)
Other way:\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(đúng)
Dấu "=" xảy ra khi a=b=c
1 ) (a+b+c)^2 >= 3(ab+bc+ac)
<=> a^2 + b^2 + c^2 >= ab + bc + ac
<=> 2a^2 + 2b^2 + 2c^2 >= 2ab + 2bc + 2ac
<=> a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + a^2 - 2ac + c^2 >= 0
<=> (a - b)^2 + (b-c)^2 + (a-c)^2 >= 0
( luôn đúng với mọi a ; b ; c )
( đpcm )
2 ) P = \(\frac{\left(a+b+c\right)^2}{ab+bc+ac}+\frac{ab+bc+ac}{\left(a+b+c\right)^2}=\frac{\left(a+b+c\right)^2}{9\left(ab+bc+ac\right)}+\frac{ab+bc+ac}{\left(a+b+c\right)^2}+\frac{8\left(a+b+c\right)^2}{9\left(ab+bc+ac\right)}\)
AD BĐT Cô - si và BĐT phụ đã cmt ở trên ta có : \(P\ge2.\frac{1}{3}+\frac{8.3.\left(ab+bc+ac\right)}{9\left(ab+bc+ac\right)}=\frac{2}{3}+\frac{8}{3}=\frac{10}{3}\)
Dấu " = " xảy ra <=> a = b = c
Khôi Bùi : theo e ý 2 có thể đơn giản hóa vấn đề bằng cách đặt ẩn phụ
đặt \(\frac{\left(a+b+c\right)^2}{ab+bc+ca}=t\left(t\ge3\right)\)
\(\Rightarrow P=t+\frac{1}{t}=\frac{t}{9}+\frac{1}{t}+\frac{8}{9}t\)
Áp dụng BĐT AM-GM ta có:
\(P\ge2.\sqrt{\frac{t}{9}.\frac{1}{t}}+\frac{8}{9}t\ge\frac{2.1}{3}+\frac{8}{9}.3=\frac{10}{3}\)
Dấu " = " xảy ra <=> a=b