Ta đi chứng minh BĐT phụ sau :

\(\sqrt[3]{4.\left(x^3+y^3\right)}\ge x+y\) với \(x,y>0\)

BĐT tương đương :

\(4.\left(a^3+b^3\right)\ge\left(a+b\right)^3\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)( đúng )

Áp dụng vào bài toán có :

\(P\ge a+b+b+c+c+a=2.\left(a+b+c\right)=2.2019=4038\)

Dấu "=" xảy ra khi \(a=b=c=673\)

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

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

Nguyễn Việt Lâm anh làm bài này giúp em với ạ

Akai Haruma giúp em bài trên với ạ

Akai Haruma cô giúp em với !!!

có cả mấy bất đẳng thức đó hả

bn viết công thức tổng quát ra cho mk vs

mk thanks

tích đi rồi ta làm

tích đi bạn

post từng câu một thôi bn nhìn mệt quá

\(\frac{b\left(2a-b\right)}{a\left(b+c\right)}+\frac{c\left(2b-c\right)}{b\left(c+a\right)}+\frac{a\left(2c-a\right)}{c\left(a+b\right)}\le\frac{3}{2}\)

\(\Leftrightarrow\left[2-\frac{b\left(2a-b\right)}{a\left(b+c\right)}\right]+\left[2-\frac{c\left(2b-c\right)}{b\left(c+a\right)}\right]+\left[2-\frac{a\left(2c-a\right)}{c\left(a+b\right)}\right]\ge\frac{9}{2}\)

\(\Leftrightarrow\frac{b^2+2ca}{a\left(b+c\right)}+\frac{c^2+2ab}{b\left(c+a\right)}+\frac{a^2+2bc}{c\left(a+b\right)}\ge\frac{9}{2}\)

Áp dụng BĐT Schwarz, ta có :

\(\frac{b^2}{a\left(b+c\right)}+\frac{c^2}{b\left(c+a\right)}+\frac{a^2}{c\left(a+b\right)}\ge\frac{\left(a+b+c\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\)( 1 )

\(\frac{ac}{a\left(b+c\right)}+\frac{ab}{b\left(c+a\right)}+\frac{bc}{c\left(a+b\right)}=\frac{c^2}{c\left(b+c\right)}+\frac{a^2}{a\left(a+c\right)}+\frac{b^2}{b\left(a+b\right)}\)           ( 2 )

\(\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+ab+bc+ac}\)

Cộng ( 1 ) với ( 2 ), ta được :

\(\frac{b^2+2ca}{a\left(b+c\right)}+\frac{c^2+2ab}{b\left(c+a\right)}+\frac{a^2+2bc}{c\left(a+b\right)}\)

\(\ge\left(a+b+c\right)^2\left(\frac{1}{2\left(ab+bc+ac\right)}+\frac{2}{a^2+b^2+c^2+ab+bc+ac}\right)\)

\(\ge\left(a+b+c\right)^2\left(\frac{\left(1+2\right)^2}{2\left(ab+bc+ac\right)+2\left(a^2+b^2+c^2+ab+bc+ac\right)}\right)=\frac{9}{2}\)

không biết cách này ổn không 

Ta có : \(\frac{b\left(2a-b\right)}{a\left(b+c\right)}=\frac{2-\frac{b}{a}}{\frac{c}{b}+1}\) ; tương tự :...

đặt \(\frac{a}{c}=x;\frac{b}{a}=y;\frac{c}{b}=z\Rightarrow xyz=1\)

\(\Sigma\frac{2-y}{z+1}\le\frac{3}{2}\)          

\(\Leftrightarrow2\Sigma xy^2+2\Sigma x^2+\Sigma xy\ge3\Sigma x+6\)( quy đồng khử mẫu )

\(\Leftrightarrow\Sigma\frac{x}{y}\ge\Sigma x\)( xyz = 1 )           ( luôn đúng )

\(\Rightarrowđpcm\)

1.\(\left(3x+1\right)\sqrt{2x^2-1}=5x^2+\frac{3}{2}x-3\)ĐK \(2x^2-1\ge0\)

<=> \(10x^2-3x-6-2\left(3x+1\right)\sqrt{2x^2-1}=0\)

<=> \(7x^2-4x-8+\left(3x+1\right)\left(x+2-2\sqrt{2x^2-1}\right)=0\)

<=>\(7x^2-4x-8+\left(3x+1\right).\frac{\left(x+2\right)^2-4\left(2x^2-1\right)}{x+2+2\sqrt{2x^2-1}}=0\)

<=> \(7x^2-4x-8+\left(3x+1\right).\frac{-7x^2+4x+8}{x+2+2\sqrt{2x^2-1}}=0\)

<=>\(\orbr{\begin{cases}7x^2-4x-8=0\left(1\right)\\1-\frac{3x+1}{x+2+2\sqrt{2x^2-1}}=0\left(2\right)\end{cases}}\)

Giải (2)

\(2\sqrt{2x^2-1}=2x-1\)

<=> \(\hept{\begin{cases}x\ge\frac{1}{2}\\4x^2+4x-5=0\end{cases}}\)

=> \(x=\frac{-1+\sqrt{6}}{2}\)(thỏa mãn ĐKXĐ)

Giải (1)=> \(x=\frac{2+2\sqrt{15}}{7}\)

Vậy \(S=\left\{\frac{2+2\sqrt{15}}{7},\frac{-1+\sqrt{6}}{2}\right\}\)

Áp dụng BĐT AM-GM ta có:

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

Tương tự ta cũng có:\(\sqrt{b\left(c+2a\right)}\le\frac{3b+c+2a}{2\sqrt{3}}\)

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

Cộng theo vế các BĐT lại ta được:

\(VT\le\frac{3a+b+2c}{2\sqrt{3}}+\frac{3b+c+2a}{2\sqrt{3}}+\frac{3c+a+2b}{2\sqrt{3}}=\frac{6a+6b+6c}{2\sqrt{3}}=\frac{6.4}{2\sqrt{3}}=4\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{4}{3}\)