Cho a,b,c là 3 số dương thỏa mãn abc = 1
Chứng minh
\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}\ge\dfrac{1}{4}\)
Cho a, b, c là các số dương biết abc = 1. Chứng minh rằng: \(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}\ge\dfrac{1}{2}\)
\(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b+1}{12}+\dfrac{c+2}{18}\ge3\sqrt[3]{\dfrac{a^3\left(b+1\right)\left(c+2\right)}{216\left(b+1\right)\left(c+2\right)}}=\dfrac{a}{2}\)
Tương tự: \(\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c+1}{12}+\dfrac{a+2}{18}\ge\dfrac{b}{2}\)
\(\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}+\dfrac{a+1}{12}+\dfrac{b+2}{18}\ge\dfrac{c}{2}\)
Cộng vế:
\(VT+\dfrac{5}{36}\left(a+b+c\right)+\dfrac{7}{12}\ge\dfrac{1}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{13}{36}\left(a+b+c\right)-\dfrac{7}{12}\ge\dfrac{13}{36}.3\sqrt[3]{abc}-\dfrac{7}{12}=\dfrac{1}{2}\) (đpcm)
Cho các số dương a,b,c cs abc=1 Chứng minh rằng
\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}\ge\dfrac{1}{4}\)
\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b+2}{36}+\dfrac{c+3}{48}\ge3\sqrt[3]{\dfrac{a^3\left(b+2\right)\left(c+3\right)}{1728\left(b+2\right)\left(c+3\right)}}=\dfrac{a}{4}\)
Tương tự: \(\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c+2}{36}+\dfrac{a+3}{48}\ge\dfrac{b}{4}\)
\(\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}+\dfrac{a+2}{36}+\dfrac{b+3}{48}\ge\dfrac{c}{4}\)
Cộng vế:
\(P+\dfrac{7\left(a+b+c\right)}{144}+\dfrac{17}{48}\ge\dfrac{a+b+c}{4}\)
\(\Rightarrow P\ge\dfrac{29}{144}\left(a+b+c\right)-\dfrac{17}{48}\ge\dfrac{29}{144}.3\sqrt[3]{abc}-\dfrac{17}{48}=\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho các số thực dương a,b,c có abc=1 chứng minh rằng:
\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}\ge\dfrac{1}{4}\)
Cho a, b, c là ba số dương thỏa mãn \(abc\)=1. Chứng minh rằng:
\(\dfrac{1}{a^3\left(b+c\right)}\)+\(\dfrac{1}{b^3\left(a+c\right)}\)+\(\dfrac{1}{c^3\left(a+b\right)}\)≥\(\dfrac{3}{2}\)
Đặt \(P=\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(abc\right)^2}{a^3\left(b+c\right)}+\dfrac{\left(abc\right)^2}{b^3\left(c+a\right)}+\dfrac{\left(abc\right)^2}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ca\right)^2}{b\left(c+a\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\)
\(P\ge\dfrac{\left(bc+ca+ab\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\) (BĐT B.C.S)
\(=\dfrac{ab+bc+ca}{2}\) \(\ge\dfrac{3\sqrt[3]{abbcca}}{2}=\dfrac{3}{2}\) (do \(abc=1\)).
ĐTXR \(\Leftrightarrow a=b=c=1\)
1.Cho \(a,b,c,d\) là các số nguyên thỏa mãn \(a^3+b^3=2\left(c^3-d^3\right)\) . Chứng minh rằng a+b+c+d chia hết cho 3
2.Cho ba số dương a,b,c thỏa mãn abc=1. Chứng minh rằng \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)
thử bài bất :D
Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)
Hoàn toàn tương tự:
\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)
\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)
Cộng (*),(**),(***) vế theo vế ta được:
\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)
Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )
Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=1
1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D
Cho 3 số thực dương x,y,z thỏa mãn \(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}=3\)
Chứng minh \(\dfrac{27a^2}{c\left(c^2+9a^2\right)}+\dfrac{b^2}{a\left(4a^2+b^2\right)}+\dfrac{8c^3}{b\left(9b^2+4c^2\right)}\ge\dfrac{3}{2}\)
cho các số dương a,b,c thỏa mãn abc=1. chứng minh rằng
\(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c^3}{\left(1+a\right)\left(1+c\right)}\ge\dfrac{3}{4}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b+1}{8}+\dfrac{c+1}{8}\)
\(\ge3\sqrt[3]{\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}\cdot\dfrac{b+1}{8}\cdot\dfrac{c+1}{8}}=\dfrac{3a}{4}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c+1}{8}+\dfrac{a+1}{8}\ge\dfrac{3b}{4};\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{a+1}{8}+\dfrac{b+1}{8}\ge\dfrac{3c}{4}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT+\dfrac{2\left(a+b+c+3\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)
\(\Leftrightarrow VT+\dfrac{2\left(3\sqrt[3]{abc}+3\right)}{8}\ge\dfrac{3\cdot3\sqrt[3]{abc}}{4}\Leftrightarrow VT\ge\dfrac{3}{4}=VP\)
Khi \(a=b=c=1\)
Với a, b, c là những số thực dương thỏa mãn \(\left(a+b\right)\left(b+c\right)\)\(\left(c+a\right)\)=1
Chứng minh rằng \(\dfrac{a}{b\left(b+2c\right)^2}\)+\(\dfrac{b}{c\left(c+2a\right)^2}\)+\(\dfrac{c}{a\left(a+2b\right)^2}\)≥\(\dfrac{4}{3}\)
cho a,b,c là các số thực dương. Chứng minh rằng :
\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(a+b+c\right)\)
AD bđt AM-GM cho 3 số
\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+C}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c}{a^3\left(b+c\right)}.\dfrac{\left(b+c\right)}{4bc}.\dfrac{1}{2b}}=\dfrac{3}{2a}\)
\(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}\ge\dfrac{3}{2a}-\dfrac{3}{4b}-\dfrac{1}{4c}\)
thiết lập bđt tương tự r cộng lại \(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\left(\dfrac{3}{2}-\dfrac{3}{4}-\dfrac{1}{4}\right)\left(a+b+c\right)=\dfrac{1}{2}\left(a+b+c\right)\)