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![](https://rs.olm.vn/images/avt/0.png?1311)
Đề bài mình sửa lại : A = a2021 - b2021 + c2021 - (a - b + c)2021
Ta có \(\sqrt{a}-\sqrt{b}+\sqrt{c}=\sqrt{a-b+c}\)
\(\Leftrightarrow a+b+c-2\sqrt{ab}-2\sqrt{bc}+2\sqrt{ca}=a-b+c\)
\(\Leftrightarrow b-\sqrt{ab}-\sqrt{bc}+\sqrt{ca}=0\)
\(\Leftrightarrow\sqrt{b}\left(\sqrt{b}-\sqrt{a}\right)-\sqrt{c}\left(\sqrt{b}-\sqrt{a}\right)=0\)
\(\Leftrightarrow\left(\sqrt{b}-\sqrt{c}\right).\left(\sqrt{b}-\sqrt{a}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=c\\b=a\end{matrix}\right.\)
Với b = c
A = a2021 - b2021 + c2021 - (a - b + c)2021
= a2021 - a2021
= 0
Tương tự với b = a ta được A = 0
Vậy A = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(a=b=c=\dfrac{2021}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\sqrt{a+b}+\sqrt{b+c}\sqrt{c+a}\)
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 ;-;)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Q\le\sqrt{3\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\le\sqrt{6.\sqrt{3\left(a^2+b^2+c^2\right)}}=\sqrt{6\sqrt{3}}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Lại có:
\(a^2+b^2+c^2\le1\Rightarrow0\le a;b;c\le1\)
\(\Leftrightarrow a\left(a-1\right)+b\left(b-1\right)+c\left(c-1\right)\le0\)
\(\Leftrightarrow a+b+c\ge a^2+b^2+c^2=1\)
Do đó:
\(Q^2=2\left(a+b+c\right)+2\sqrt{a^2+ab+bc+ca}+2\sqrt{b^2+ab+bc+ca}+2\sqrt{c^2+ab+bc+ca}\)
\(Q^2\ge2\left(a+b+c\right)+2\sqrt{a^2}+2\sqrt{b^2}+2\sqrt{c^2}\)
\(Q^2\ge4\left(a+b+c\right)\ge4\)
\(\Rightarrow Q\ge2\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)
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 \(Max\left(P\right)=\sqrt{12126}\Leftrightarrow a=b=c=\frac{2021}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(3a+1\ge\left(\dfrac{\sqrt{10}-1}{3}a+1\right)^2\Leftrightarrow a\left(3-a\right)\ge0\) (luôn đúng)
Do đó \(\sqrt{3a+1}\ge\dfrac{\sqrt{10}-1}{3}a+1\).
Tương tự, \(\sqrt{3b+1}\ge\dfrac{\sqrt{10}-1}{3}b+1;\sqrt{3c+1}\ge\dfrac{\sqrt{10}-1}{3}c+1\).
Do đó \(\sqrt{3a+1}+\sqrt{3b+1}+\sqrt{3c+1}\ge\sqrt{10}+2\).
Dấu "=" xảy ra khi chẳng hạn a = 3; b = c = 0
Tham khảo:
https://hoc24.vn/hoi-dap/tim-kiem?id=219071991005&q=Cho%203%20s%E1%BB%91%20th%E1%BB%B1c%20kh%C3%B4ng%20%C3%A2m%20a%2Cb%2Cc%20v%C3%A0%20a%20b%20c%3D3%20T%C3%ACm%20GTLN%20v%C3%A0%20GTNN%20c%E1%BB%A7a%20bi%E1%BB%83u%20th%E1%BB%A9c%20K%3D%5C%28%5Csqrt%7B3a%201%7D%20%5Csqrt%7B3b%201%7D%20%5Csqrt%7B3c%201%7D%5C%29
BĐT Bunhiacopski:
\(P^2\le3\left(2a+2b+2c\right)=6.2021=12126\)
\(\Leftrightarrow P\le\sqrt{12126}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{2021}{3}\)