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\(Q=ac+bc-2022ab\le ac+bc=c\left(a+b\right)\le\dfrac{1}{4}\left(c+a+b\right)^2=\dfrac{1}{4}\)
\(Q_{max}=\dfrac{1}{4}\) khi \(\left\{{}\begin{matrix}a+b+c=1\\ab=0\\c=a+b\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(0;\dfrac{1}{2};\dfrac{1}{2}\right);\left(\dfrac{1}{2};0;\dfrac{1}{2}\right)\)
\(Q=c\left(a+b\right)-2022ab\ge c\left(a+b\right)-\dfrac{1011}{2}\left(a+b\right)^2\)
\(Q\ge c\left(1-c\right)-\dfrac{1011}{2}\left(1-c\right)^2\)
\(Q\ge c\left(1-c\right)-\dfrac{1011}{2}c\left(c-2\right)-\dfrac{1011}{2}\)
\(Q\ge\dfrac{c\left(1011+1013\left(1-c\right)\right)}{2}-\dfrac{1011}{2}\ge-\dfrac{1011}{2}\)
\(Q_{min}=-\dfrac{1011}{2}\) khi \(\left(a;b;c\right)=\left(\dfrac{1}{2};\dfrac{1}{2};0\right)\)
1
Áp dụng BĐT Cauchy cho 2 số dương:
4ac=2.b.2c≤2(b+2c2)2≤2(a+b+2c2)2=2.(12)2=12
⇒−4bc≥−12
⇒K=ab+4ac−4bc≥−4bc≥−12
Áp dụng BĐT Bunhiacopxki ta có:
\(\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)\)
\(=3\left(2a+2b+2c\right)=3.2\left(a+b+c\right)=6.2021=12126\)
\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{12126}\)
Dấu ''='' xảy ra khi \(a=b=c=\dfrac{2021}{3}\)
Lời giải:
Đặt $a+b+c=p; ab+bc+ac=q=1; abc=r$
$p,r\geq 0$
Áp dụng BĐT AM-GM: $p^2\geq 3q=3\Rightarrow p\geq \sqrt{3}$
$a,b,c\leq 1\Leftrightarrow (a-1)(b-1)(c-1)\leq 0$
$\Leftrightarrow p+r\leq 2\Rightarrow p\leq 2$
$P=\frac{(a+b+c)^2-2(ab+bc+ac)+3}{a+b+c-abc}=\frac{(a+b+c)^2+1}{a+b+c-abc}=\frac{p^2+1}{p-r}$
Ta sẽ cm $P\geq \frac{5}{2}$ hay $P_{\min}=\frac{5}{2}$
$\Leftrightarrow \frac{p^2+1}{p-r}\geq \frac{5}{2}$
$\Leftrightarrow 2p^2-5p+2+5r\geq 0(*)$
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Thật vậy:
Áp dụng BĐT Schur thì:
$p^3+9r\geq 4p\Rightarrow 5r\geq \frac{20}{9}p-\frac{5}{9}p^3$
Khi đó:
$2p^2-5p+2+5r\geq 2p^2-5p+2+\frac{20}{9}p-\frac{5}{9}p^3=\frac{1}{9}(2-p)(5p^2-8p+9)\geq 0$ do $p\leq 2$ và $p\geq \sqrt{3}$
$\Rightarrow (*)$ được CM
$\Rightarrow P_{\min}=\frac{5}{2}$
Dấu "=" xảy ra khi $(a,b,c)=(1,1,0)$ và hoán vị
Do \(a^2+b^2+c^2=1\Rightarrow0\le a;b;c\le1\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-1\right)\left(b-1\right)\left(c-1\right)\le0\\b^{2011}\le b\\c^{2011}\le c\end{matrix}\right.\)
\(\Rightarrow T\le a+b+c-ab-bc-ca=\left(a-1\right)\left(b-1\right)\left(c-1\right)+1-abc\le1-abc\le1\)
\(T_{max}=1\) khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và các hoán vị
Lời giải:
Do $b\leq c; a^2\geq 0$ nên $a^2(b-c)\leq 0$
$\Rightarrow Q\leq b^2(c-b)+c^2(1-c)$
Áp dụng BĐT AM-GM:
\(b^2(c-b)=4.\frac{b}{2}.\frac{b}{2}(c-b)\leq 4\left(\frac{\frac{b}{2}+\frac{b}{2}+c-b}{3}\right)^3=\frac{4}{27}c^3\)
\(\Rightarrow Q\leq c^2-\frac{23}{27}c^3=c^2(1-\frac{23}{27}c)=(\frac{54}{23})^2.\frac{23}{54}c.\frac{23}{54}c(1-\frac{23}{27}c)\leq (\frac{54}{23})^2\left(\frac{\frac{23}{54}c+\frac{23}{54}c+1-\frac{23}{27}c}{3}\right)^3=\frac{108}{529}\)
Vậy $Q_{max}=\frac{108}{529}$
Giá trị này đạt tại $(a,b,c)=(0,\frac{12}{23}, \frac{18}{23})$
Lời giải:
Do $b\leq c; a^2\geq 0$ nên $a^2(b-c)\leq 0$
$\Rightarrow Q\leq b^2(c-b)+c^2(1-c)$
Áp dụng BĐT AM-GM:
\(b^2(c-b)=4.\frac{b}{2}.\frac{b}{2}(c-b)\leq 4\left(\frac{\frac{b}{2}+\frac{b}{2}+c-b}{3}\right)^3=\frac{4}{27}c^3\)
\(\Rightarrow Q\leq c^2-\frac{23}{27}c^3=c^2(1-\frac{23}{27}c)=(\frac{54}{23})^2.\frac{23}{54}c.\frac{23}{54}c(1-\frac{23}{27}c)\leq (\frac{54}{23})^2\left(\frac{\frac{23}{54}c+\frac{23}{54}c+1-\frac{23}{27}c}{3}\right)^3=\frac{108}{529}\)
Vậy $Q_{max}=\frac{108}{529}$
Giá trị này đạt tại $(a,b,c)=(0,\frac{12}{23}, \frac{18}{23})$
\(P=\sqrt{a+b}+\sqrt{b+c}\sqrt{c+a}\)
Aps dụng Bunhia-cốpxki : \(P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)\)
\(=6\left(a+b+c\right)\)
\(=6.2021=12126\Leftrightarrow P=\sqrt{12126}\)
Vậy \(Max\left(P\right)=\sqrt{12126}\Leftrightarrow a=b=c=\dfrac{2021}{3}\)
(Refer ;-;)
\(P=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)
áp dụng bunhia - cốpxki
\(P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)\)
\(=6\left(a+b+c\right)\)
\(=6.2021=12126< =>P=\sqrt{12126}\)
vậy MAX P=\(\sqrt{12126}\)
\(P=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)
\(\Rightarrow P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)
Áp dụng BĐT Bunyakovsky ta có:
\(P^2\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)=6\left(a+b+c\right)=6\cdot2021\)
\(\Rightarrow P\le\sqrt{6\cdot2021}=\sqrt{12126}\)
Dấu "=" xảy ra khi: \(a=b=c=\frac{2021}{3}\)
Vậy \(Max\left(P\right)=\sqrt{12126}\Leftrightarrow a=b=c=\frac{2021}{3}\)