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Do \(-1\le a;b;c\le1\)

\(\Rightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)+\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge0\)

\(\Leftrightarrow1-abc-a-b-c+ab+bc+ca+1+abc+b+c+c+ab+bc+ca\ge0\)

\(\Leftrightarrow2\left(ab+bc+ca\right)+2\ge0\)

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

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

\(\Leftrightarrow a^2+b^2+c^2\le2\)

Mà \(\left|a\right|;\left|b\right|;\left|c\right|\le1\Rightarrow\left\{{}\begin{matrix}a^4\le a^2\\b^6\le b^2\\c^8\le c^2\end{matrix}\right.\)

\(\Rightarrow a^4+b^6+c^8\le a^2+b^2+c^2\le2\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(-1;0;1\right)\) và các hoán vị

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)

là -2 đấy

\(-1\le a\le2\Rightarrow\hept{\begin{cases}a+1\ge0\\a-2\le0\end{cases}\Rightarrow\left(a+1\right)\left(a-2\right)\le0}\)

Tương tự \(\left(b+1\right)\left(b-2\right)\le0,\left(c+1\right)\left(c-2\right)\le0\)

=> (a+1)(a-2)+(b+1)(b-2)+(c+1)(c-2)\(\le\)0 => a²+b²+c²-(a+b+c)-6\(\le\)0 

=>a²+b²+c² \(\le\)6 

Dấu "=" xảy ra <=> (a+1)(  a-2)=0, (b+1)(b-2)=0, (c+1)(c-2)=0 , a+b+c=0 <=> a=2, b=c=-1 và các hoán vị 

Do \(\left\{{}\begin{matrix}a\ge0\\b\ge1\\a+b+c=5\end{matrix}\right.\) \(\Rightarrow c\le4\)

\(\Rightarrow2\le c\le4\Rightarrow\left(c-2\right)\left(c-4\right)\le0\Rightarrow c^2\le6c-8\)

\(0\le a\le1< 6\Rightarrow a\left(a-6\right)\le0\Rightarrow a^2\le6a\)

\(1\le b\le2< 5\Rightarrow\left(b-1\right)\left(b-5\right)\le0\Rightarrow b^2\le6b-5\)

Cộng vế:

\(a^2+b^2+c^2\le6\left(a+b+c\right)-13=17\)

\(A_{max}=17\) khi \(\left(a;b;c\right)=\left(0;1;4\right)\)

Giải:

Vì \(0\leq a,b,c\leq 1\Rightarrow ab,ac,ab\geq abc\)

Do đó mà \(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\leq \frac{a+b+c}{abc+1}\)

Giờ chỉ cần chỉ ra \(\frac{a+b+c}{abc+1}\leq 2\). Thật vậy:

Do \(0\leq b,c\leq 1\Rightarrow (b-1)(c-1)\geq 0\Leftrightarrow bc+1\geq b+c\Rightarrow bc+a+1\geq a+b+c\)

Suy ra \( \frac{a+b+c}{abc+1}\leq \frac{bc+a+1}{abc+1}=\frac{bc+a-2abc-1}{abc+1}+2=\frac{(bc-1)(1-a)-abc}{abc+1}+2\)

Ta có \(\left\{\begin{matrix}bc\le1\\a\le1\\abc\ge0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}\left(bc-1\right)\left(1-a\right)\le1\\-abc\le0\end{matrix}\right.\) \(\Rightarrow \frac{(bc-1)(1-a)-abc}{abc+1}+2\leq 2\Rightarrow \frac{a+b+c}{abc+1}\leq 2\)

Chứng minh hoàn tất

Dấu bằng xảy ra khi \((a,b,c)=(0,1,1)\) và hoán vị.

vao cau hoi hay OLM itm