Chắc áp dụng được Cauchy-Schwarz

Ta có: \(\sqrt[3]{\left(a+b\right).\frac{2}{3}.\frac{2}{3}}\le\frac{a+b+\frac{4}{3}}{3}=\frac{a+b}{3}+\frac{4}{9}\)

Tương tự rồi cộng các vế của BĐT lại, ta được: \(\sqrt[3]{\frac{4}{9}}P\le\frac{2\left(a+b+c\right)}{3}+\frac{4}{3}=2\Rightarrow P\le\sqrt[3]{18}\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)

lỡ k Kiệt Nguyễn r, bài Kiệt Nguyên sai r

\(Q\le\sqrt{3\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\le\sqrt{6.\sqrt{3\left(a^2+b^2+c^2\right)}}=\sqrt{6\sqrt{3}}\)

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

cái cuối là \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\)  nha

\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)

\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Tương tự:

\(\dfrac{1}{\sqrt{b^2-bc+c^2}}\le\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\le\dfrac{1}{2}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)

Cộng vế:

\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(\sqrt{\dfrac{ab}{c+ab}}=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)

Tương tự: \(\sqrt{\dfrac{bc}{a+bc}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\) ; \(\sqrt{\dfrac{ca}{b+ca}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)\)

Cộng vế với vế:

\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{b}{a+b}+\dfrac{a}{a+b}\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

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 \(a=b=c=\dfrac{2021}{3}\)

\(P\le\sqrt{3\left(9a+16b+9b+16c+9c+16a\right)}=\sqrt{75\left(a+b+c\right)}=15\)

\(P_{max}=15\) khi \(a=b=c=1\)

\(6a+3b+2c=abc\Leftrightarrow\dfrac{2}{ab}+\dfrac{3}{ac}+\dfrac{6}{bc}=1\)

Đặt \(\left(\dfrac{1}{a};\dfrac{2}{b};\dfrac{3}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)

\(Q=\dfrac{1}{\sqrt{\dfrac{1}{x^2}+1}}+\dfrac{2}{\sqrt{\dfrac{4}{y^2}+4}}+\dfrac{3}{\sqrt{\dfrac{9}{z^2}+9}}=\dfrac{x}{\sqrt{x^2+1}}+\dfrac{y}{\sqrt{y^2+1}}+\dfrac{z}{\sqrt{z^2+1}}\)

\(Q=\dfrac{x}{\sqrt{x^2+xy+yz+zx}}+\dfrac{y}{\sqrt{y^2+xy+yz+zx}}+\dfrac{z}{\sqrt{z^2+xy+yz+zx}}\)

\(Q=\dfrac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\dfrac{y}{\sqrt{\left(x+y\right)\left(y+z\right)}}+\dfrac{z}{\sqrt{\left(x+z\right)\left(y+z\right)}}\)

\(Q\le\dfrac{1}{2}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)=\dfrac{3}{2}\)

\(Q_{max}=\dfrac{3}{2}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\) hay \(\left(a;b;c\right)=\left(\sqrt{3};2\sqrt{3};3\sqrt{3}\right)\)

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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.