Question

[Tôi Là Ai]

cho \(0\le a\le2,0\le b\le2,0\le c\le2\) và a+b+c=3.chứng minh \(a^2+b^2+c^2\le5\)

Answer 1

[Thắng Nguyễn]

\(\left(2-a\right)\left(2-b\right)\left(2-c\right)\ge0\)

\(\Leftrightarrow8-4\left(a+b+c\right)+2\left(ab+bc+ca\right)-abc\ge0\)

\(\Leftrightarrow2\left(ab+bc+ca\right)\ge4\left(a+b+c\right)-8+abc\)

\(\Leftrightarrow2\left(ab+bc+ca\right)\ge12-8+abc\ge4\)

\(\Rightarrow2\left(ab+bc+ca\right)\ge4\)

\(\Rightarrow-2\left(ab+bc+ca\right)\le-4\)

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^2+b^2+c^2=9-2\left(ab+bc+ca\right)\le9-4=5\)(Đpcm)

Dấu = khi \(\hept{\begin{cases}\left(2-a\right)\left(2-b\right)\left(2-c\right)=0\\abc=0\\a+b+c=3\end{cases}}\)

\(\Rightarrow\left(a;b;c\right)=\left(2;1;0\right)\)và hoán vị.

Answer 2

[Namikaze Minato]

a = 2 ( t/m )

b = 1 ( t/m )

c = 0 ( t/m )

vậy \(a^2+b^2+c^2\le5\)