có cả mấy bất đẳng thức đó hả

bn viết công thức tổng quát ra cho mk vs

mk thanks

Ta có: \(a+b+c+\sqrt{abc}=4\)

\(\Rightarrow4a+4b+4c+4\sqrt{abc}=16\)

\(\Rightarrow4a+4\sqrt{abc}=16-4b-4c\)

\(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(16-4b-4c+bc\right)}=\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\)

\(=\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}=\left|2a+\sqrt{abc}\right|=2a+\sqrt{abc}\)

Tương tự: 

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\\\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\end{matrix}\right.\)

\(\Rightarrow A=\sqrt{a\left(4-b\right)\left(4-c\right)}+\sqrt{b\left(4-c\right)\left(4-a\right)}+\sqrt{c\left(4-a\right)\left(4-b\right)}-\sqrt{abc}=2a+2b+2c+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c+\sqrt{abc}\right)=8\)

Ta có \(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(a+c+\sqrt{abc}\right)\left(4-c\right)}\)

\(=\sqrt{\left(a^2+ac+a\sqrt{abc}\right)\left(4-c\right)}\\ =\sqrt{4a^2+ac\left(4-\sqrt{abc}-a-c\right)+4a\sqrt{abc}}\\ =\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}\\ =2a+\sqrt{abc}\left(a,b,c>0\right)\)

Cmtt \(\sqrt{b\left(4-c\right)\left(4-a\right)}=2b+\sqrt{abc};\sqrt{c\left(4-b\right)\left(4-a\right)}=2c+\sqrt{abc}\)

\(\Rightarrow A=2\left(a+b+c\right)+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c\right)+2\sqrt{abc}\\ A=2\left(a+b+c+\sqrt{abc}\right)=2\cdot4=8\)

post từng câu một thôi bn nhìn mệt quá

Bài 1.

Ta có:\(\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)=x^2+2020-x^2=2020\)

\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)\)

\(\Rightarrow y+\sqrt{y^2+2020}=\sqrt{x^2+2020}-x\)

\(\Rightarrow x+y=\sqrt{x^2+2020}-\sqrt{y^2+2020}\)   (1)

Ta có:\(\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)=y^2+2020-y^2=2020\)

\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)\)

\(\Rightarrow x+\sqrt{x^2+2020}=\sqrt{y^2+2020}-y\)

\(\Rightarrow x+y=\sqrt{y^2+2020}-\sqrt{x^2+2020}\)          (2)

Cộng vế với vế của (1) và (2) ta có:

\(2\left(x+y\right)=\sqrt{y^2+2020}-\sqrt{x^2+2020}+\sqrt{x^2+2020}-\sqrt{y^2+2020}\)

\(\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0\)

Bài 2: 

Ta có: (2a+1)(2b+1)=9

nên \(2b+1=\dfrac{9}{2a+1}\)

\(\Leftrightarrow2b=\dfrac{9}{2a+1}-\dfrac{2a+1}{2a+1}=\dfrac{8-2a}{2a+1}\)

\(\Leftrightarrow b=\dfrac{8-2a}{4a+2}=\dfrac{4-a}{2a+1}\)

\(\Leftrightarrow b+2=\dfrac{4-a+4a+2}{2a+1}=\dfrac{3a+6}{2a+1}\)

Ta có: \(A=\dfrac{1}{a+2}+\dfrac{1}{b+2}\)

\(=\dfrac{1}{a+2}+\dfrac{2a+1}{3a+6}\)

\(=\dfrac{3+2a+1}{3a+6}\)

\(=\dfrac{2a+4}{3a+6}=\dfrac{2}{3}\)

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 \(a=b=c=\dfrac{3}{\sqrt{6063}}\)

Nè bạn :) 

Ta có : \(2ab+2ac\ge4a\sqrt{bc}\) (Cauchy_)

\(\Rightarrow a^2+2ab+2ac+4bc\ge a^2+4a\sqrt{bc}+4bc\)

\(\Rightarrow a^2+2ab+2ac+4bc\ge\left(a+2\sqrt{bc}\right)^2\)

\(\Rightarrow\sqrt{\left(a+2b\right)\left(a+2c\right)}\ge a+2\sqrt{bc}\)\(\left(1\right)\)

Tương tự : \(\sqrt{\left(b+2a\right)\left(b+2c\right)}\ge b+2\sqrt{ac}\)\(\left(2\right)\)

\(\sqrt{\left(c+2a\right)\left(c+2b\right)}\ge c+2\sqrt{ab}\)\(\left(3\right)\)

Từ \(\left(1\right);\left(2\right);\left(3\right)\)\(\Rightarrow\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge3\)

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

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

Thay vào biểu thức M ta được M = \(\frac{\sqrt{3}}{3}\)