Ta chứng minh bất đẳng thức sau  

Với x, y, z > 0 ta luôn có  \(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\)  (1)

Theo BĐT Cô-si

\(x^4+x^4+y^4+z^4\ge4\sqrt[4]{x^8y^4z^4}=4x^2yz\)

\(y^4+y^4+z^4+x^4\ge4\sqrt[4]{y^8z^4x^4}=4y^2zx\)

\(z^4+z^4+x^4+y^4\ge4\sqrt[4]{z^8x^4y^4}=4z^2xy\)

Cộng vế theo vế ta được:  \(4\left(x^4+y^4+z^4\right)\ge4\left(x^2yz+y^2zx+z^2xy\right)\)

\(\Leftrightarrow\)  \(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\)

Vậy (1) đc c/m

Bất đẳng thức cần c/m có thể viết lại thành

\(\frac{abcd}{a^4+b^4+c^4+abcd}+\frac{abcd}{b^4+c^4+d^4+abcd}+\frac{abcd}{c^4+d^4+a^4+abcd}+\frac{abcd}{d^4+a^4+b^4+abcd}\le1\)

Áp dụng (1) ta có  

\(\frac{abcd}{a^4+b^4+c^4+abcd}\le\frac{abcd}{abc\left(a+b+c\right)+abcd}=\frac{abcd}{abc\left(a+b+c+d\right)}=\frac{d}{a+b+c+d}\)

Tương tự  

\(\frac{abcd}{b^4+c^4+d^4+abcd}\le\frac{a}{a+b+c+d}\)

\(\frac{abcd}{c^4+d^4+a^4+abcd}\le\frac{b}{a+b+c+d}\)

\(\frac{abcd}{d^4+a^4+b^4+abcd}\le\frac{c}{a+b+c+d}\)

Cộng theo vế suy ra đpcm.

\(1-\frac{a}{a+1}=\frac{1}{1+a}=\frac{c}{c+1}+\frac{b}{b+1}+\frac{d}{d+1}\Rightarrow\frac{1}{a+1}\ge3\sqrt[3]{\frac{bcd}{\left(b+1\right)\left(c+1\right)\left(d+1\right)}}\)

cmtt rồi nhân 3 cái lại vs nhau => đpcm

☘ Áp dụng bất đẳng thức AM - GM

\(\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}+\dfrac{d}{1+d}=1\)

\(\Leftrightarrow1-\dfrac{a}{1+a}=\dfrac{b}{1+b}+\dfrac{c}{1+c}+\dfrac{d}{1+d}\)

\(\Rightarrow\dfrac{1}{1+a}\ge3\sqrt[3]{\dfrac{bcd}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)

☘ Tương tự, ta cũng có:

\(\dfrac{1}{1+b}\ge3\sqrt[3]{\dfrac{acd}{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)

\(\dfrac{1}{1+c}\ge3\sqrt[3]{\dfrac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)

\(\dfrac{1}{1+d}\ge3\sqrt[3]{\dfrac{abc}{\left(1+a\right)\left(1+c\right)\left(1+b\right)}}\)

☘ Nhân vế theo vế

\(\Rightarrow\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge\dfrac{81abcd}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\)

\(\Rightarrow abcd\le\dfrac{1}{81}\)

☘ Dấu "=" xảy ra khi \(a=c=b=d=\dfrac{1}{3}\)

⚠ Nguồn: https://hoc24.vn/hoi-dap/question/463672.html

Đường link : Câu hỏi của Hà Lê - Toán lớp 9 - Học toán với OnlineMath

Ta có : a⁴ + b⁴ \(\ge\)2a²b² ; b⁴ + c⁴ \(\ge\)2b²c² ; a⁴ + c⁴ \(\ge\)2a²c²

\(\Rightarrow\)a⁴ + b⁴ + c⁴ \(\ge\)a²b² + b²c² + a²c² ( 1 )

Lại có : a²b² + b²c² \(\ge\)2b²ac ; b²c² + a²c² \(\ge\)2c²ab ; a²b² + a²c² \(\ge\)2a²bc

\(\Rightarrow\)a²b² + b²c² + a²c² \(\ge\)abc ( a + b + c ) ( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\)a⁴ + b⁴ + c⁴ \(\ge\) abc ( a + b + c ) 

Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c = 1

Tương tự , b⁴ + c⁴ + d⁴ ​​​\(\ge\)​bcd ( b + c + d ) ; a⁴ + b⁴ + d⁴ ​\(\ge\)​abd ( a + b + d ) ; c⁴ + d⁴ + a⁴ ​\(\ge\)​acd ( a + c + d ) 

\(\frac{1}{a^4+b^4+c^4+abcd}\le\frac{1}{abc\left(a+b+c\right)+abcd}=\frac{abcd}{abc\left(a+b+c+d\right)}=\frac{d}{a+b+c+d}\)

\(\frac{1}{b^4+c^4+d^4+abcd}\le\frac{a}{a+b+c+d}\); \(\frac{1}{a^4+b^4+d^4+abcd}\le\frac{c}{a+b+c+d}\)

\(\frac{1}{c^4+d^4+a^4+abcd}\le\frac{b}{a+b+c+d}\)

Cộng từng vế theo vế , ta được : 

A \(\le\)1  ( đặt A = biểu thức ấy nhé )

Vậy GTLN A = 1 \(\Leftrightarrow\)a = b = c = d = 1

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

Tương tự: \(\dfrac{1}{\left(1+c\right)^2}+\dfrac{1}{\left(1+d\right)^2}\ge\dfrac{1}{1+cd}\)

\(\Rightarrow B\ge\dfrac{1}{1+ab}+\dfrac{1}{1+cd}=\dfrac{1}{1+ab}+\dfrac{1}{1+\dfrac{1}{ab}}=\dfrac{1}{1+ab}+\dfrac{ab}{1+ab}=1\)

\(B_{min}=1\) khi \(a=b=c=d=1\)

Áp dụng BĐT phụ ta có:

\(B\ge\dfrac{1}{1+ab}+\dfrac{1}{1+cd}=\dfrac{ab+cd+2}{1+ab+cd+abcd}=1\)

Vậy GTNN của B bằng 1 <=> a=b=c=d=1

A = 1/(a + 1) + 1/(b + 1) + 1/(c + 1) + 1/(d + 1) ≥ 3 
→ 1/(a + 1) ≥ 1 - 1/(b + 1) + 1 - 1/(c + 1) + 1 - 1/(d + 1) 
→ 1/(a + 1) ≥ b/(b + 1) + c/(c + 1) + d/(d + 1) 
áp dụng BĐT Cauchy cho 3 số dương: 
b/(b + 1) + c/(c + 1) + d/(d + 1) ≥ 3 ³√(bcd)/[(b + 1)(c + 1)(d + 1)] 
→ 1/(a + 1) ≥ 3 ³√(bcd)/[(b + 1)(c + 1)(d + 1)] tương tự 
1/(b + 1) ≥ 3 ³√(acd)/[(a + 1)(c + 1)(d + 1)] 
1/(c + 1) ≥ 3 ³√(abd)/[(a + 1)(b + 1)(d + 1)] 
1/(d + 1) ≥ 3 ³√(abc)/[(a + 1)(b + 1)(c + 1)] 
nhân theo vế → 1/[(a + 1)(b + 1)(c + 1)(d + 1)] ≥ 81abcd/[(a + 1)(b + 1)(c + 1)(d + 1)] 
→ 1 ≥ 81abcd → abcd ≤ 1/81 

Ẹt số xui đưa link cũng bị duyệt

Áp dụng BĐT AM-GM ta có: 

\(\frac{1}{d+1}=1-\frac{d}{d+1}\ge\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\)

\(\ge3\sqrt[3]{\frac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\). TƯơng tự cho 3 BĐT còn lại

\(\frac{1}{a+1}\ge3\sqrt[3]{\frac{bcd}{\left(b+1\right)\left(c+1\right)\left(d+1\right)}};\frac{1}{b+1}\ge3\sqrt[3]{\frac{acd}{\left(a+1\right)\left(c+1\right)\left(d+1\right)}};\frac{1}{c+1}\ge3\sqrt[3]{\frac{abd}{\left(a+1\right)\left(b+1\right)\left(d+1\right)}}\)

Nhân theo vế 4 BDT trên ta có: 

\(\frac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\ge81\sqrt[3]{\left(\frac{abcd}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\right)^3}\)

\(\Leftrightarrow\frac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\ge\frac{81abcd}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\)

Hay ta có ĐPCM

Áp dụng BĐT AM-GM ta có:

\(\dfrac{1}{a+1}\ge1-\dfrac{1}{b+1}+1-\dfrac{1}{c+1}+1-\dfrac{1}{d+1}\)

\(=\dfrac{b}{b+1}+\dfrac{c}{c+1}+\dfrac{d}{d+1}\)\(\ge3\sqrt[3]{\dfrac{bcd}{\left(b+1\right)\left(c+1\right)\left(d+1\right)}}\)

Tương tự cho 3 BĐT còn lại cũng có:

\(\dfrac{1}{1+b}\ge3\sqrt[3]{\dfrac{acd}{\left(a+1\right)\left(c+1\right)\left(d+1\right)}};\dfrac{1}{c+1}\ge3\sqrt[3]{\dfrac{abd}{\left(a+1\right)\left(b+1\right)\left(d+1\right)}};\dfrac{1}{d+1}\ge3\sqrt[3]{\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)

Nhân theo vế 4 BĐT trên ta có:

\(\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\ge81\sqrt[3]{\left(\dfrac{abcd}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\right)^3}\)

\(\Leftrightarrow1\ge81abcd\Leftrightarrow abcd\le\dfrac{1}{81}\)

Từ giả thiết, ta có:

\(\dfrac{1}{1+a}\ge1-\dfrac{1}{1+b}+1-\dfrac{1}{1+c}+1-\dfrac{1}{1+d}=\dfrac{b}{1+b}+\dfrac{c}{c+1}+\dfrac{d}{d+1}\ge3\sqrt[3]{\dfrac{b.c.d}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)

Tương tự:

\(\dfrac{1}{1+b}\ge3\sqrt[3]{\dfrac{cda}{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)

\(\dfrac{1}{1+c}\ge3\sqrt[3]{\dfrac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)

\(\dfrac{1}{1+d}\ge3\sqrt[3]{\dfrac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

Nhân vế theo vế 4 BĐT vừa chứng minh rồi rút gọn ta được:

\(abcd\le\dfrac{1}{81}\left(đpcm\right)\)

Mỗi vế trừ đi 4