Ta có: BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)( CM bằng BĐT Shwars nha).Áp dụng ta có:

\(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5a}+\frac{1}{3a+2b+4c}\ge\frac{9}{9a+6b+12c}=\frac{3}{3a+2b+4c}\left(1\right)\)

\(\frac{1}{b+3c+5a}+\frac{1}{c+3a+5b}+\frac{1}{3b+2c+4a}\ge\frac{9}{9b+6c+12a}=\frac{3}{3b+2c+4a}\left(2\right)\)

\(\frac{1}{c+3a+5b}+\frac{1}{a+3b+5c}+\frac{1}{3c+2a+4b}\ge\frac{9}{9c+6a+12b}=\frac{3}{3c+2a+4b}\left(3\right)\)

Cộng (1),(2) và (3) có:

\(2\left(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5c}+\frac{1}{c+3a+5b}\right)+\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\ge3\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\)

\(\Rightarrow2VP\ge2VT\)

\(\RightarrowĐPCM\)

b+1+2c chứ ko phải b+1=2c nhé

A=1

lợi dụng a+b+c=1 thay vào từng mẫu

Nhầm, cái cuối là \(\frac{4}{4+d+2ad+3abd}\)

\(\frac{1}{1+2a+3ab+4abc}+\frac{2}{2+3b+4bc+bcd}+\frac{3}{3+4c+cd+2acd}+\frac{4}{4+d+2ad+3abd}\)

= \(\frac{1}{1+2a+3ab+4abc}+\frac{2a}{2a+3ab+4abc+abcd}+\frac{3ab}{3ab+4abc+abcd+2abacd}\)

\(+\frac{4abc}{4abc+abcd+2aabcd+3abcabd}\)

= \(\frac{1}{1+2a+3ab+4abc}+\frac{2a}{2a+3ab+4abc+1}+\frac{3ab}{3ab+4abc+1+2a}+\frac{4abc}{4abc+1+2a+3ab}\)

= \(\frac{1+2a+3ab+4abc}{1+2a+3ab+4abc}=1\)

Vì  abcd=1 nên : a=1 ;b=1;c=1;d=1

       thay số vào pt ta đc : \(\frac{1}{1+2\cdot1+3\cdot1\cdot1+4\cdot1\cdot1}\)+ \(\frac{1}{2+3\cdot1+4\cdot1\cdot1+1\cdot1\cdot1}\)+ \(\frac{1}{3+4\cdot1+1\cdot1+2\cdot1\cdot1\cdot1}\)+ \(\frac{1}{4+1+2\cdot1\cdot1+3\cdot1\cdot1\cdot1}\)

                    Tương đương : \(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)= \(\frac{4}{10}\)=\(\frac{2}{5}\)

a , b , c , d cũng có thể âm mà Long

Ta có: \(a^2-b^2=4c^2\)

\(\Rightarrow a^2-b^2-4c^2=0\)

Xét hiệu:

 \(\left(5a-3b+8c\right)\left(5a-3b-8c\right)-\left(3a-5b\right)^2\)

\(=\left(5a-3b\right)^2-\left(8c\right)^2-\left(3a-5b\right)^2\)

\(=25a^2-30ab+9b^2-64c^2-9a^2+30ab-25b^2\)

\(=16a^2-16b^2-64c^2\)

\(=16\left(a^2-b^2-4c^2\right)\)

\(=16.0\)

\(=0\)

\(\Rightarrow\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)

                                                                             đpcm 

Tham khảo nhé~

Một cách khác :))

Xét VT của biểu thức cần cm ta có :

( 5a - 3b + 8c )( 5a - 3b - 8c )

= [ ( 5a - 3b ) + 8c ][ ( 5a - 3b ) - 8c ]

= ( 5a - 3b )² - ( 8c )²

= 25a² - 30ab + 9b² - 64c²

= 25a² - 30ab + 9b² - 16.4c²

= 25a² - 30ab + 9b² - 16( a² - b² ) < theo đề a² - b² = 4c² >

= 25² - 30ab + 9b² - 16a² + 16b²

= 9a² - 30ab + 25b²

= ( 3a - 5b )² = VP

=> đpcm