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 \(a=b=c=\dfrac{2021}{3}\)

bn sử dụng bất đẳng thức cô si đi

Nguyễn Đại Nghĩa,bác nói cụ thể hơn được ko :v

\(P=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)

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 \(Max\left(P\right)=\sqrt{12126}\Leftrightarrow a=b=c=\frac{2021}{3}\)

Ta có: \(P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)

\(=2\left(a+b+c\right)+2\left[\sqrt{\left(a+b\right)\left(b+c\right)}+\sqrt{\left(b+c\right)\left(c+a\right)}+\sqrt{\left(c+a\right)\left(a+b\right)}\right]\)

\(=4042+2\left[\sqrt{\left(a+b\right)\left(b+c\right)}+\sqrt{\left(b+c\right)\left(c+a\right)}+\sqrt{\left(c+a\right)\left(a+b\right)}\right]\)

Mà \(\left(a+b\right)\left(b+c\right)\ge\left(0+b\right)\left(b+0\right)=b^2\)

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

\(\Rightarrow P\ge4042+2\left(a+b+c\right)=4042+4042=8084\)

\(\Rightarrow P\ge2\sqrt{2021}\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}a=2021\\b=c=0\end{cases}}\) và các hoán vị của nó

Vậy \(Min\left(P\right)=2\sqrt{2021}\Leftrightarrow\hept{\begin{cases}a=2021\\b=c=0\end{cases}}\)

\(P=\sqrt{a+b}+\sqrt{b+c}\sqrt{c+a}\)

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 ;-;)

Đề thi học kỳ 1 trường Ams

**Min

Từ \(a^2+b^2+c^2=1\Rightarrow a^2\le1;b^2\le1;c^2\le1\)

\(\Rightarrow a\le1;b\le1;c\le1\Rightarrow a^2\le a;b^2\le b;c^2\le c\)

Khi đó:

\(\sqrt{a+b^2}\ge\sqrt{a^2+b^2};\sqrt{b+c^2}\ge\sqrt{b^2+c^2};\sqrt{c+a^2}\ge\sqrt{c^2+a^2}\)

\(\Rightarrow P\ge\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\)

\(\Rightarrow P\ge\sqrt{1-c^2}+\sqrt{1-a^2}+\sqrt{1-b^2}\)

Ta có:

\(\sqrt{1-c^2}\ge1-c^2\Leftrightarrow1-c^2\ge1-2c^2+c^4\Leftrightarrow c^2\left(1-c^2\right)\ge0\left(true!!!\right)\)

Tương tự cộng lại:

\(P\ge3-\left(a^2+b^2+c^2\right)=2\)

dấu "=" xảy ra tại \(a=b=0;c=1\) and hoán vị.

**Max

Có BĐT phụ sau:\(\sqrt{a}+\sqrt{b}+\sqrt{c}\le\sqrt{3\left(a+b+c\right)}\left(ezprove\right)\)

Áp dụng:

\(\sqrt{a+b^2}+\sqrt{b+c^2}+\sqrt{c+a^2}\)

\(\le\sqrt{3\left(a+b+c+a^2+b^2+c^2\right)}\)

\(=\sqrt{3\left(a+b+c\right)+3}\)

\(\le\sqrt{3\left(\sqrt{3\left(a^2+b^2+c^2\right)}+3\right)}=\sqrt{3\cdot\sqrt{3}+3}\)

Dấu "=" xảy ra tại \(a=b=c=\pm\frac{1}{\sqrt{3}}\)

1≥a=>a≥a²=>24a+25= 4a+20a+25≥4a²+2.2a.5+25=(2a+5)²
=>\(\sqrt{24a+25}\)≥2a+5
cmtt=> K≥ 2(a+b+c)+15=17
dấu "=" xảy ra  <=> (a,b,c)~(1,0,0)

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xin lỗi mk nhầm bài

Ta có:

\(\sqrt{12a+\left(b-c\right)^2}=\sqrt{4a\left(a+b+c\right)+\left(b-c\right)^2}\)

\(=\sqrt{4a^2+4ab+4ac+b^2-2bc+c^2}\)

\(=\sqrt{\left(2a+b+c\right)^2-4bc}\)

\(\le\sqrt{\left(2a+b+c\right)^2}=2a+b+c\)

Khi đó \(K\le4\left(a+b+c\right)=12\)

Dấu "=" xảy ra tại \(a=0;b=0;c=3\) và các hoán vị.