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Lời giải :
Ta có: \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}\ge3\)
\(\Leftrightarrow\frac{1}{1+a}\ge1-\frac{1}{1+b}+1-\frac{1}{1+c}+1-\frac{1}{1+d}\)
\(\Leftrightarrow\frac{1}{1+a}\ge\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\ge3\sqrt[3]{\frac{bcd}{\left(b+1\right)\left(c+1\right)\left(d+1\right)}}\) ( Cô-si )
Chứng minh tương tự ta cũng có :
\(\frac{1}{1+b}\ge3\sqrt[3]{\frac{acd}{\left(a+1\right)\left(c+1\right)\left(d+1\right)}}\); \(\frac{1}{1+c}\ge3\sqrt[3]{\frac{abd}{\left(a+1\right)\left(b+1\right)\left(d+1\right)}}\);
\(\frac{1}{1+d}\ge3\sqrt[3]{\frac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
Nhân theo vế 4 BĐT ta được :
\(\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge81\sqrt[3]{\frac{a^3b^3c^3d^3}{\left(a+1\right)^3\left(b+1\right)^3\left(c+1\right)^3\left(d+1\right)^3}}\)
\(\Leftrightarrow\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge81\cdot\frac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\)
\(\Leftrightarrow1\ge81\cdot abcd\)
\(\Leftrightarrow abcd\le\frac{1}{81}\)
Ta có đpcm.
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=d=\frac{1}{3}\)
chết dòng thứ 5 từ dưới lên thiếu biến \(d\) trên tử số :( ai rủ lòng thương sửa hộ phát :>
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)
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)
Còn đề bạn viết sai rồi nhé
Đặt: \(P=\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}\)
Ta có:
\(\frac{a+1}{b^2+1}=a-\frac{ab^2-1}{b^2+1}\ge a-\frac{ab^2-1}{2b}=a-\frac{ab}{2}+\frac{1}{2b}\)
Tương tự ta có:
\(\frac{b+1}{c^2+1}\ge b-\frac{bc}{2}+\frac{1}{2c},\frac{c+1}{a^2+1}\ge c-\frac{ca}{2}+\frac{1}{2a}\)
\(\Rightarrow P\ge a+b+c-\frac{ab+bc+ca}{2}+\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3-\frac{\left(a+b+c\right)^2}{6}+\frac{1}{2}\left(\frac{\left(1+1+1\right)^2}{a+b+c}\right)\)
\(=3-\frac{9}{6}+\frac{1}{2}.\frac{9}{3}=3\)
Dấu bằng xảy ra khi a=b=c=1
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:
\(\frac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)}\ge81\sqrt[3]{\left(\frac{abcd}{\left(a+1\right)\left(b+1\right)\left(d+1\right)}\right)^3}\)
\(\Rightarrow abcd\le\frac{1}{81}\)
\(1-\frac{a}{a+1}=\frac{1}{1+a}=\frac{c}{c+1}+\frac{b}{b+1}+\frac{d}{d+1}\Rightarrow\frac{1}{a+1}\ge3\sqrt[3]{\frac{bcd}{\left(b+1\right)\left(c+1\right)\left(d+1\right)}}\)
cmtt rồi nhân 3 cái lại vs nhau => đpcm
Theo BĐT AM-GM: \(a^4+b^4\ge2a^2b^2\)
Tương tự suy ra \(a^4+b^4+c^4\)\(\ge a^2b^2+b^2c^2+c^2a^2\)
Tiếp tục dùng AM-GM: \(a^2b^2+b^2c^2=b^2\left(a^2+c^2\right)\ge2ab^2c\)
Tương tự suy ra \(a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\)
\(\Rightarrow a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
\(\Rightarrow a^4+b^4+c^4+abcd\ge abc\left(a+b+c\right)+abcd\)\(=abc\left(a+b+c+d\right)\)
\(\Rightarrow\frac{1}{a^4+b^4+c^4+abcd}\le\frac{1}{abc\left(a+b+c+d\right)}\)
Tương tự cho 3 BĐT còn lại rồi cộng theo vế:
\(VT\le\frac{a+b+c+d}{abcd\left(a+b+c+d\right)}=\frac{1}{abcd}=VP\)
\(a+b+c+ab+bc+ca=6abc\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)
Đặt \(\left\{{}\begin{matrix}\frac{1}{a}=x\\\frac{1}{b}=y\\\frac{1}{c}=z\end{matrix}\right.\)\(\Rightarrow x+y+z+xy+yz+zx=6\)
CM \(P=x^2+y^2+z^2\ge3\)
\(x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z\)
\(2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)
Cộng vế với vế
\(\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)=12\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge9\)
\(\Rightarrow x^2+y^2+z^2\ge3\left(đpcm\right)\)
Vậy dấu "=" xảy ra khi \(x=y=z=1\) hoặc \(a=b=c=1\)
\(a+b+c+ab+bc+ca=6abc\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)
Ta lại có:
\(\frac{1}{a^2}+1+\frac{1}{b^2}+1+\frac{1}{c^2}+1-3\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}-3\)
\(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}\ge\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)
Cộng vế với vế:
\(\frac{3}{a^2}+\frac{3}{b^2}+\frac{3}{c^2}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)-3\)
\(\Leftrightarrow3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge6.2-3=9\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
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: \(1\ge81abcd\Rightarrow abcd\le\frac{1}{81}\)
