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1/\(=4a^2+4b^2+c^2+8ab-4bc-4ca+4b^2+4c^2+a^2+8bc-4ca-4ab+4a^2+4c^2+b^2+8ca-4bc-4ab=\)
\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)
2/
Ta có
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge-2\left(ab+bc+ca\right)=2\)
\(\Rightarrow P=9\left(a^2+b^2+c^2\right)\ge18\)
\(\Rightarrow P_{min}=18\)
Ta có: \(a^2+1=a^2+ab+bc+ca=\left(a+b\right)\left(c+a\right)\)
Tương tự: \(\left\{{}\begin{matrix}b^2+1=\left(a+b\right)\left(b+c\right)\\c^2+1=\left(c+a\right)\left(b+c\right)\end{matrix}\right.\)
=> \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)
Mặt khác: \(a+b+c-abc=a\left(1-bc\right)+b+c\)
\(=a\left(ab+ca\right)+b+c\) (Vì ab+bc+ca=1)
\(=\left(a^2+1\right)\left(b+c\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (Vì \(a^2+1=\left(a+b\right)\left(c+a\right)\))
\(T=1\)
\(P=\dfrac{1}{bc\left(b+c\right)+2023}+\dfrac{1}{ca\left(c+a\right)+2023}+\dfrac{1}{ab\left(a+b\right)+2023}\left(abc=2023\right)\)
\(\Leftrightarrow P=\dfrac{1}{bc\left(b+c\right)+abc}+\dfrac{1}{ca\left(c+a\right)+abc}+\dfrac{1}{ab\left(a+b\right)+abc}\)
\(\Leftrightarrow P=\dfrac{1}{bc\left(a+b+c\right)}+\dfrac{1}{ca\left(a+b+c\right)}+\dfrac{1}{ab\left(a+b+c\right)}\)
\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\)
\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left[\dfrac{a^2bc+b^2ca+c^2ab}{\left(abc\right)^2}\right]\)
\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left[\dfrac{abc\left(a+b+c\right)}{\left(abc\right)^2}\right]\)
\(\Leftrightarrow P=\dfrac{1}{abc}=\dfrac{1}{2023}\)
làm cái đề ra ấy, ngại viết lại đề :P
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=4\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)
\(\Rightarrow M=1^{2018}+1^{2019}+1^{2020}=1+1+1=3\)
Dễ dàng chứng minh được:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)với \(x,y>0\)(1)
Dấu bằng xảy ra \(\Leftrightarrow x=y>0\)
Ta có:
\(\frac{a}{bc\left(a+1\right)}=\frac{a}{abc+bc}=\frac{a}{ab+bc+ca+bc}=\frac{a}{\left(ab+bc\right)+\left(bc+ca\right)}\)
Áp dụng (1), ta được:
\(\frac{1}{ab+bc}+\frac{1}{bc+ca}\ge\frac{4}{\left(ab+bc\right)+\left(bc+ca\right)}\)
\(\Leftrightarrow\frac{1}{4\left(ab+bc\right)}+\frac{1}{4\left(bc+ca\right)}\ge\frac{1}{ab+bc+bc+ca}\)
\(\Leftrightarrow\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ca}\right)\ge\frac{a}{ab+bc+bc+ca}\)
\(\Leftrightarrow\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ca}\right)\ge\frac{a}{bc\left(a+1\right)}\left(2\right)\)
Dấu bằng xảy ra \(\Leftrightarrow b=c>0\)
Chúng minh tương tự, ta được:
\(\frac{b}{4}\left(\frac{1}{ab+ca}+\frac{1}{bc+ca}\right)\ge\frac{b}{ca\left(b+1\right)}\left(3\right)\)
Dấu bằng xảu ra \(\Leftrightarrow a=c>0\).
\(\frac{c}{4}\left(\frac{1}{ac+ab}+\frac{1}{ab+bc}\right)\ge\frac{c}{ab\left(c+1\right)}\left(4\right)\)
Từ (2), (3) và (4), ta được:
\(\frac{a}{bc\left(a+1\right)}+\frac{b}{ca\left(b+1\right)}+\frac{c}{ab\left(c+1\right)}\le\)\(\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ac}\right)+\frac{b}{4}\left(\frac{1}{ac+bc}+\frac{1}{ac+ab}\right)\)\(+\frac{c}{4}\left(\frac{1}{ab+bc}+\frac{1}{ab+ac}\right)\)
\(\Leftrightarrow P\le\frac{1}{4}.\left(\frac{a}{ab+bc}+\frac{c}{ab+bc}\right)+\frac{1}{4}\left(\frac{a}{bc+ac}+\frac{b}{bc+ac}\right)\)\(+\frac{1}{4}\left(\frac{b}{ab+ac}+\frac{c}{ab+ac}\right)\)
\(\Leftrightarrow P\le\frac{a+c}{4\left(ab+bc\right)}+\frac{a+b}{4\left(bc+ac\right)}+\frac{b+c}{4\left(ab+ac\right)}\)
\(\Leftrightarrow P\le\frac{a+c}{4b\left(a+c\right)}+\frac{a+b}{4c\left(a+b\right)}+\frac{b+c}{4a\left(b+c\right)}\)
\(\Leftrightarrow P\le\frac{1}{4b}+\frac{1}{4c}+\frac{1}{4a}\)
\(\Leftrightarrow P\le\frac{1}{4}\left(\frac{ab+bc+ca}{abc}\right)\)
\(\Leftrightarrow P\le\frac{1}{4}.\frac{abc}{abc}=\frac{1}{4}.1=\frac{1}{4}\)( vì \(ab+bc+ca=abc\))
Dấu bằng xảy ra
\(\Leftrightarrow\hept{\begin{cases}a=b=c>0\\ab+bc+ca=abc\end{cases}}\Leftrightarrow a=b=c=3\)
Vậy \(minP=\frac{1}{4}\Leftrightarrow a=b=c=3\)
Ta có: \(3\left(a^2+b^2+c^2\right)=\left(a+b+c\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow2ab+2bc+2ac=2\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)(1)
Mà \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Rightarrow\left(1\right)\)xảy ra \(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Leftrightarrow a=b=c\)
\(\Rightarrow M=ab+bc+ca-\left(a+b+c\right)+1=3a^2-3a+1\)
\(=\left(\sqrt{3}a\right)^2-2.\sqrt{3}a.\frac{\sqrt{3}}{2}+\frac{3}{4}+\frac{1}{4}\)
\(=\left(\sqrt{3}a-\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)
(Dấu "=" \(\Leftrightarrow\sqrt{3}a-\frac{\sqrt{3}}{2}=0\Leftrightarrow a=\frac{1}{2}\)
hay \(a=b=c=\frac{1}{2}\)
Vậy \(M_{min}=\frac{1}{4}\Leftrightarrow a=b=c=\frac{1}{2}\)
giả thiết \(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\) (biến đổi tương đương)
Thay xuống: \(M=3a^2-3a+1=3\left(a-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)
Đẳng thức xảy ra khi \(a=\frac{1}{2}\)
P/s; hướng làm là đưa về 1 biến như vậy đó, khi tính toán có thể có sai số, bạn tự check lại.
\(a^2+b^2+c^2=ab+bc+ca\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Rightarrow\left(2a^2+2b^2+2c^2\right)-\left(2ab+2bc+2ca\right)=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\)\(\Rightarrow a-b=b-c=c-a=0\)
\(\Rightarrow P=\left(a-b\right)^{2015}+\left(b-c\right)^{2016}+\left(c-a\right)^{2017}=0\)
cảm ơn bạn nha