\(-b\le a\le b\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b\ge0\\a-b\le0\end{matrix}\right.\)

\(a^2-b^2=\left(a-b\right)\cdot\left(a+b\right)\le0\)

\(\Leftrightarrow a^2\le b^2\)

BĐT chỉ đúng khi \(b\ge0\) (Dễ thấy nếu b < 0 thì -b > 0 > b, bđt sai)

Bình phương 2 vế em nhé, GTTĐ bình phương thì âm hay dương nó cx như nhau

\(\left|a+b\right|\le\left|a\right|+\left|b\right|\)

\(\Leftrightarrow\left(a+b\right)^2\le\left(\left|a\right|+\left|b\right|\right)^2\)

\(\Leftrightarrow a^2+2ab+b^2\le a^2+2\left|ab\right|+b^2\)

Áp dụng BĐT cosi:

\(a\sqrt{1-b^2}=\sqrt{a^2\left(1-b^2\right)}\le\dfrac{a^2+1-b^2}{2}\)

Tương tự cx có: \(b\sqrt{1-c^2}\le\dfrac{b^2+1-c^2}{2}\)

\(c\sqrt{1-a^2}\le\dfrac{c^2+1-a^2}{2}\)

Cộng vế với vế \(\Rightarrow VT\le\dfrac{3}{2}\)

Dấu = xảy ra <=> \(\left\{{}\begin{matrix}a^2=1-b^2\\b^2=1-c^2\\c^2=1-a^2\end{matrix}\right.\) \(\Leftrightarrow a^2+b^2+c^2=3-\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow a^2+b^2+c^2=\dfrac{3}{2}\) (đpcm)

Bạn tham khảo ở đây nhé

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Nếu a,b nguyên ms đc

\(a^2+b^2\ge2ab\)( bồ đề dễ dàng CM)

\(\Rightarrow2ab\le a^2+b^2\le8\)

\(\Rightarrow a^2+b^2+2ab\le16\Leftrightarrow\left(a+b\right)^2\le16\)

\(\Rightarrow-4\le a+b\le4\)

Vì \(0\le a\le2;0\le b\le2;0\le c\le2\Rightarrow\left(2-a\right)\left(2-b\right)\left(2-c\right)\ge0\)\(\Leftrightarrow8-4\left(a+b+c\right)+2\left(ab+bc+ca\right)-abc\ge0\)\(\Leftrightarrow2\left(ab+bc+ca\right)\ge4\left(a+b+c\right)-8+abc\ge4\)\(\Leftrightarrow2\left(ab+bc+ca\right)\ge12-8+abc\ge4\)

\(\Rightarrow\)\(2\left(ab+bc+ca\right)\ge4\)

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

Ta có :

\(a+b+c=3\Rightarrow\left(a+b+c\right)^2=9\)

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

\(\Rightarrow a^2+b^2+c^2=9-2\left(ab+bc+ca\right)\le9-4=5\Rightarrowđpcm\)Đẳng thức xảy ra khi

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

\(\left[{}\begin{matrix}2-a=0\\2-b=0\\2-c=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)