ở đây Một bài toán siêu khó

Ta có:

\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}\ge3\)

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

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

Áp dụng BĐT Cauchy cho 3 số dương:

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

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

Tương tự:

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

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

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

Nhân theo vế ta được:

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

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

Vậy \(abcd\le\frac{1}{81}\) (Đpcm)

Từ giả thiết  => \(\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\le1-\frac{a}{a+1}=\frac{1}{a+1}\)

Áp dụng bđt Cauchy cho 3 số dương : \(\frac{1}{a+1}\ge\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\ge3.\sqrt[3]{\frac{bcd}{\left(b+1\right)\left(c+1\right)\left(d+1\right)}}\). Tương tự: \(\frac{1}{b+1}\ge3.\sqrt[3]{\frac{acd}{\left(a+1\right)\left(c+1\right)\left(d+1\right)}}\)

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

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

Nhân từ 4 bđt: \(1\ge81abcd\Rightarrow abcd\le\frac{1}{81}\)

Bài 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 

TK NHA

Áp dụng BDT AM-GM ta có:

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

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

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

Tương tự cho các BĐT còn lại cũng có:

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

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

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

Nhân theo vế 4 BĐT trên ta có:

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

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

Ta có: \(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\le1\)

\(\Rightarrow\left\{\begin{matrix}\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\le1-\frac{d}{d+1}=\frac{1}{d+1}\\\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\le1-\frac{a}{a+1}=\frac{1}{a+1}\\\frac{a}{a+1}+\frac{c}{c+1}+\frac{d}{d+1}\le1-\frac{b}{b+1}=\frac{1}{b+1}\\\frac{a}{a+1}+\frac{b}{b+1}+\frac{d}{d+1}\le1-\frac{c}{c+1}=\frac{1}{c+1}\end{matrix}\right.\)

Áp dụng BĐT Cauchy cho 3 bộ số thực không âm

\(\Rightarrow\left\{\begin{matrix}\frac{1}{d+1}\ge\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\ge3\sqrt[3]{\frac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\\\frac{1}{a+1}\ge\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\ge3\sqrt[3]{\frac{bcd}{\left(b+1\right)\left(c+1\right)\left(d+1\right)}}\\\frac{1}{b+1}\ge\frac{a}{a+1}+\frac{c}{c+1}+\frac{d}{d+1}\ge3\sqrt[3]{\frac{acd}{\left(a+1\right)\left(c+1\right)\left(d+1\right)}}\\\frac{1}{c+1}\ge\frac{a}{a+1}+\frac{b}{b+1}+\frac{d}{d+1}\ge3\sqrt[3]{\frac{abd}{\left(a+1\right)\left(b+1\right)\left(d+1\right)}}\end{matrix}\right.\)

Nhân từng vế:

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

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

\(\Rightarrow1\ge81abcd\)

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

p/s : lí do tớ tự trả lời câu hỏi của mình là để coi câu trả lời của mình có đúng hay ko thôi nha , mong các bạn đứng có hiểu lầm , nếu bạn nào có cách nào nhanh và gọn hơn thì phiền các bạn chỉ dùm luôn nha.

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

https://olm.vn/hoi-dap/detail/223126660207.html?pos=512235459592

Giờ mình mới để ý , câu này có trong chuyên đề : Bất đẳng thức Cauchy (Cô si) của cô Nguyễn Linh Chi (ở phần dạng toán và hướng dẫn giải) (mình đã inbox link cho bạn rồi)

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