\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)

\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)

\(\sqrt{ab}+\sqrt{4b.c}+2\left(a+c\right)\le\dfrac{1}{2}\left(a+b\right)+\dfrac{1}{2}\left(4b+c\right)+2\left(a+c\right)=\dfrac{5}{2}\left(a+b+c\right)\)

\(\Rightarrow P\ge\dfrac{2}{5}\left(\dfrac{1}{a+b+c}-\dfrac{1}{\sqrt{a+b+c}}\right)=\dfrac{2}{5}\left(\dfrac{1}{\sqrt{a+b+c}}-\dfrac{1}{2}\right)^2-\dfrac{1}{10}\ge-\dfrac{1}{10}\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}a+b+c=4\\a=b=\dfrac{c}{4}\end{matrix}\right.\) em tự giải ra a;b;c

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

\(P=\sqrt{\dfrac{yz}{x^2+1}}+\sqrt{\dfrac{zx}{y^2+1}}+\sqrt{\dfrac{xy}{z^2+1}}\)

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

\(P=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\dfrac{zx}{\left(y+z\right)\left(x+y\right)}}+\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\)

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

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

ko biết mk làm có đúng ko nhma có gì sai thì đừng trách mk nhé

\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\ge\dfrac{63}{a^2+b^2+c^2}\)

\(6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{a}{ac}\right)+2021\ge\dfrac{54}{ab+bc+ac}+2021\ge\dfrac{54}{a^2+b^2+c^2}+2021\)

<=>\(\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{2021}{9}\)

\(p^2=\left(\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\right)^2\)

áp dụng bđt \(a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)

\(p^2\le3.\left(\dfrac{1}{3\left(2a^2+b^2\right)}+\dfrac{1}{3\left(2b^2+c^2\right)}+\dfrac{1}{3\left(2c^2+a^2\right)}\right)=\dfrac{1}{2a^2+b^2}+\dfrac{1}{2b^2+c^2}+\dfrac{1}{2c^2+a^2}\)

\(< =>p^2\le\dfrac{9}{2a^2+b^2+2b^2+c^2+2c^2+a^2}\)

<=> \(p^2\le3.\dfrac{1}{a^2+b^2+c^2}=\dfrac{2021}{3}< =>p\le\sqrt{\dfrac{2021}{3}}\)

dấu bằng xảy ra khi \(a=b=c=\sqrt{\dfrac{3}{2021}}\)

\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)+2021\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2021\)

\(\Rightarrow2021\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\sqrt{2021.3}=\sqrt{6063}\)

Từ đó:

\(\sqrt{3\left(2a^2+b\right)}=\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}\ge\sqrt{\left(2a+b\right)^2}=2a+b\)

\(\Rightarrow\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}\le\dfrac{1}{2a+b}=\dfrac{1}{a+a+b}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)\)

Tương tự: \(\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{c}+\dfrac{1}{a}\right)\)

Cộng vế:

\(\Rightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{\sqrt{6063}}{3}\)

\(P_{max}=\dfrac{\sqrt{6063}}{3}\) khi \(a=b=c=\dfrac{3}{\sqrt{6063}}\)

Ta có \(\sqrt{bc\left(1+a^2\right)}=\sqrt{bc+a^2bc}=\sqrt{bc+a\left(a+b+c\right)}\)

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

Đặt BT đề cho là P

\(\Leftrightarrow P=\sum\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}=\sum\sqrt{\dfrac{a}{a+b}\cdot\dfrac{a}{a+c}}\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{b+a}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\cdot3=\dfrac{3}{2}\)

Dấu \("="\Leftrightarrow a=b=c=\sqrt{3}\)

\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)

\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)

\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)