giúp tôi vs, tôi đang cần gấp

tra mạng ko có ???

ko có:(

Ta có

   \(a+b+c=6\)

  \(\Leftrightarrow\left(a+b+c\right)^2=36\)

  \(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=36\)

   Mà \(a^2+b^2+c^2=ab+bc+ca\)

 Khi đó ta có

     \(3\left(ab+bc+ca\right)=36\)

 \(\Leftrightarrow ab+bc+ca=12\)

  \(\Leftrightarrow\hept{\begin{cases}2ab+2bc+2ca=24\\2a^2+2b^2+2c^2=24\end{cases}}\)

 \(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}}\Leftrightarrow a=b=c=\frac{6}{3}=2\)  ( 1 )

  Thay (1) vào C ta có

        \(C=\left(1-2\right)^{2021}+\left(2-1\right)^{2021}+\left(2-2\right)^{2021}\)

             \(=-1+1+0=0\)

         Vậy ......................

a² + b² + c² = ab + bc + ca 

<=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0

<=> (a² - 2ab + b²) + (b² - 2ca + c²) + (c² - 2ac + a²) = 0

<=> (a - b)² + (b - c)² + (c - a)² = 0

Dễ thấy   (a - b)² + (b - c)² + (c - a)² \(\ge0\forall a,b,c\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow a=b=c\)

Mà a + b + c = 2025 

nên \(a=b=c=675\)

a2 + b2 + c2 = ab + bc + ca 

<=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0

<=> (a2 - 2ab + b2) + (b2 - 2ca + c2) + (c2 - 2ac + a2) = 0

<=> (a - b)2 + (b - c)2 + (c - a)2 = 0

Dễ thấy   (a - b)2 + (b - c)2 + (c - a)2 $\ge 0\forall a,b,c$

Dấu "=" xảy ra khi {�−�=0�−�=0�−�=0⇔�=�=�⎩⎨⎧​a−b=0b−c=0a−c=0​⇔a=b=c

Mà a + b + c = 2025 

nên $a=b=c=675$

\(a\text{) }\)Áp dụng: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) (a, b > 0). Dấu "=" xảy ra khi a = b.

\(\frac{1}{a^2+b^2}+\frac{1}{ab}=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}\ge\frac{4}{a^2+b^2+2ab}+\frac{1}{2.\frac{\left(a+b\right)^2}{4}}=\frac{6}{\left(a+b\right)^2}\)

\(=6\left[\frac{1}{\left(a+b\right)^2}+\frac{27}{8}\left(a+b\right)+\frac{27}{8}\left(a+b\right)\right]-\frac{81}{2}\left(a+b\right)\)

\(\ge6.3\sqrt[3]{\frac{1}{\left(a+b\right)^2}.\frac{27}{8}\left(a+b\right).\frac{27}{8}\left(a+b\right)}-\frac{81}{2}\left(a+b\right)\)

\(=\frac{81}{2}-\frac{81}{2}\left(a+b\right)\)

Tương tự: \(\frac{1}{b^2+c^2}+\frac{1}{bc}\ge\frac{81}{2}-\frac{81}{2}\left(b+c\right)\)

\(\frac{1}{c^2+a^2}+\frac{1}{ca}\ge\frac{81}{2}-\frac{81}{2}\left(c+a\right)\)

Cộng theo vế ta được 

\(A\ge3.\frac{81}{2}-81\left(a+b+c\right)=3.\frac{81}{2}-81=\frac{81}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}.\)

Vậy GTNN của A là \(\frac{81}{2}.\)

\(a^2+b^2+c^2-ab-bc-ac=0\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\) (1)

Mà: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\) 

Nên PT (1) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)

=> a = b = c

\(P=\left(a-b\right)^{2020}+\left(b-c\right)^{2021}+\left(c-a\right)^{2022}\)

\(=\left(a-a\right)^{2020}+\left(b-b\right)^{2021}+\left(c-c\right)^{2022}\)

= 0

Ta có : \(a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)

\(T=\frac{a^{2021}+b^{2021}+c^{2021}}{\left(a+b+c\right)^{2021}}=\frac{b^{2021}+b^{2021}+b^{2021}}{\left(b+b+b\right)^{2021}}=\frac{3b^{2021}}{\left(3b\right)^{2021}}=\frac{3}{3^{2021}}=\frac{1}{3^{2020}}\)