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\(DPCM\Leftrightarrow P=a^2\left(b-c\right)+b^2\left(c-b\right)+c^2\left(1-c\right)\le\frac{108}{529}\)
Ta có: \(0\le a\le b\le c\le1\Rightarrow a^2\left(b-c\right)\le0\left(1\right)\)
\(b^2\left(c-b\right)=4.\frac{b}{2}.\frac{b}{2}.\left(c-b\right)\le4\left(\frac{\frac{b}{2}+\frac{b}{2}+c-b}{3}\right)^3=\frac{4c^3}{27}\)
\(\Rightarrow P\le\frac{4c^3}{27}+c^2\left(1-c\right)=c^2\left(1-\frac{23c}{27}\right)=\frac{23c}{54}.\frac{23c}{54}\left(1-\frac{23c}{27}\right).\frac{54^2}{23^2}\)
Tiếp
\(\le\left(\frac{\frac{23c}{54}+\frac{23c}{54}+1-\frac{23c}{27}}{3}\right)^3.\frac{54^2}{23^2}=\frac{1}{27}.\frac{54^2}{23^2}=\frac{108}{529}\)
Dấu bằng xảy ra\(\Leftrightarrow\hept{\begin{cases}a^2\left(b-c\right)=0\\\frac{b}{2}=c-b\\\frac{23c}{54}=1-\frac{23c}{27}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=0\\b=\frac{2}{3}c\\c=\frac{18}{23}\end{cases}}\)
Áp dụng bđt Bunhiacopski ta có
\(\sqrt{c}.\sqrt{a-c}+\sqrt{c}.\sqrt{b-c}\le\sqrt{\left(\sqrt{c}\right)^2+\left(\sqrt{b-c}\right)^2}+\sqrt{\left(\sqrt{c}\right)^2+\left(\sqrt{a-c}\right)^2}.\)
\(\Leftrightarrow\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{c+b-c}.\sqrt{c+a-c}=\sqrt{ab}\left(đpcm\right)\)
Bu-nhi-a-cốp-ski: (ab+cd)2 \(\le\)( a2 + c2 )( b2 + d2 ) mà bạn.
\(-1\le a\le2\Rightarrow\hept{\begin{cases}a+1\ge0\\a-2\le0\end{cases}\Rightarrow\left(a+1\right)\left(a-2\right)\le0}\)
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
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)