\(ab\cdot\sqrt{\dfrac{a}{3b}}-a^2\sqrt{\dfrac{3b}{a}}\)

\(=a\sqrt{ab}-a^2\cdot\dfrac{\sqrt{3b}}{\sqrt{a}}\)

\(=a\sqrt{ab}-a\sqrt{a}\cdot\sqrt{3b}\)

\(=a\sqrt{ab}\left(1-\sqrt{3}\right)\)

\(\Leftrightarrow m=\dfrac{a\sqrt{ab}\left(1-\sqrt{3}\right)}{\sqrt{3ab}}=\dfrac{a\left(\sqrt{3}-3\right)}{3}\)

\(M\ge\dfrac{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}}{2}+\dfrac{\sqrt{\left(\sqrt{b}+\sqrt{c}\right)^2}}{2}+\dfrac{\sqrt{\left(\sqrt{c}+\sqrt{a}\right)^2}}{2}\)

\(M\ge\sqrt{a}+\sqrt{b}+\sqrt{c}=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(=\left(1^2+4^2\right)\left(a^2+\dfrac{1}{b^2}\right)\ge\left(1a+4.\dfrac{1}{b}\right)^2\\ \Rightarrow\sqrt{a^2+\dfrac{1}{vb^2}}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\) 

Tương tự

\(\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\\ \sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\\ Do.đó:\\ Q\ge\dfrac{1}{\sqrt{17}}\left(a+b+c+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)\ge\dfrac{1}{\sqrt{17}}\\ \left(a+b+c+\dfrac{36}{a+b+c}\right)\) 

\(=\dfrac{1}{\sqrt{17}}\\ \left[a+b+c+\dfrac{9}{4\left(a+b+c\right)}+\dfrac{135}{4\left(a+b+c\right)}\right]\\ \ge\dfrac{3\sqrt{17}}{2}\)

Cái thứ nhất là tại sao có cái đầu tiên =)) cái thứ 2 dấu bằng xảy ra khi nào :V

a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\)

\(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}-2}\right)\cdot\dfrac{a-4}{\sqrt{4a}}\)

\(=\dfrac{2\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{2a}\)

\(=\sqrt{a}+2\)

b: A-2<0

=>\(\sqrt{a}+2-2< 0\)

=>\(\sqrt{a}< 0\)

=>\(a\in\varnothing\)

c: Bạn ghi đầy đủ đề đi bạn

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

\(P=-\dfrac{3}{5}\) sao suy ra đc \(\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\) thế

\(A=\dfrac{1}{a}+\dfrac{1}{b}-\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)=\dfrac{1-a+b}{b}+\dfrac{1-b+a}{a}\)

Vì \(a^2+b^2=1\) và \(a,b>0\Leftrightarrow0< a< 1;0< b< 1\Leftrightarrow1+a-b>0;1-b+a>0\)

\(\Leftrightarrow A\ge2\sqrt{\dfrac{\left(1-a+b\right)\left(1-b+a\right)}{ab}}=2\sqrt{\dfrac{1-a^2-b^2+2ab}{ab}}=2\sqrt{2}\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\dfrac{1-a+b}{b}=\dfrac{1-b+a}{a}\end{matrix}\right.\Leftrightarrow a=b=\dfrac{1}{\sqrt{2}}\)

Đây là bài IMO 2001 và không cần điều kiện \(a+b+c=1\)

Áp dụng Holder:

\(P.P.\left[a\left(a^2+8bc\right)+b\left(b^2+8ac\right)+c\left(c^2+8ab\right)\right]\ge\left(a+b+c\right)^3\)

\(\Leftrightarrow P^2\ge\dfrac{\left(a+b+c\right)^3}{a^3+b^3+c^3+24abc}=\dfrac{a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)}{a^3+b^3+c^3+24abc}\)

\(\Rightarrow P^2\ge\dfrac{a^3+b^3+c^3+3.2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}}{a^3+b^3+c^3+24abc}=1\)

\(\Rightarrow P\ge1\)