Cho abc=1 ,a,b,c>0
CMR \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(a+c\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)
các bạn làm được ý nào thì làm ý đó nha
1. Cho a,b,c là độ dài 3 cạnh tam giác. Chứng minh:
a) \(\frac{1}{\left(a+b-c\right)^2}+\frac{1}{\left(a-b+c\right)^2}+\frac{1}{\left(b+c-a\right)^2}\ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
b) \(\frac{1}{\left(a+b-c\right)^3}+\frac{1}{\left(a-b+c\right)^3}+\frac{1}{\left(b+c-a\right)^3}\ge\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\)
c) \(\frac{1}{\left(a+b-c\right)^{200}}+\frac{1}{\left(a-b+c\right)^{200}}+\frac{1}{\left(b+c-a\right)^{200}}\ge\frac{1}{a^{200}}+\frac{1}{b^{200}}+\frac{1}{c^{200}}\)
d) \(\frac{1}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\sqrt{abc\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)}\)
e) \(a+b+c< \sqrt{a\left(b+c\right)}+\sqrt{b\left(a+c\right)}+\sqrt{c\left(a+b\right)}\)
f) \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}< \sqrt{6}\)
g) \(\sqrt{-a+b+c}+\sqrt{a-b+c}+\sqrt{a+b-c}\le\sqrt{3\left(a+b+c\right)}\)
cho a,b,c>0
Cm: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}=\frac{c^2}{c^2(a+b)}+\frac{a^2}{a^2(b+c)}+\frac{b^2}{b^2(c+a)}+\frac{(\sqrt[3]{abc})^2}{2abc}\)
\(\geq \frac{(c+a+b+\sqrt[3]{abc})^2}{c^2(a+b)+a^2(b+c)+b^2(c+a)+2abc}=\frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c$
1) Cho a, b, c > 0. Chứng minh: \(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
2) Cho \(a,b,c\in R\).
a) Chứng minh: \(\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right)\ge4\left(a+b+c+1\right)^2\)
b) Chứng minh: \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{16}\left(a+b+c+1\right)^2\)
3) Cho \(a,b,c\in R\)Chứng minh: \(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
2) Theo nguyên lí Dirichlet, trong ba số \(a^2-1;b^2-1;c^2-1\) có ít nhất hai số nằm cùng phía với 1.
Giả sử đó là a2 - 1 và b2 - 1. Khi đó \(\left(a^2-1\right)\left(b^2-1\right)\ge0\Leftrightarrow a^2b^2-a^2-b^2+1\ge0\)
\(\Rightarrow a^2b^2+3a^2+3b^2+9\ge4a^2+4b^2+8\)
\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\ge4\left(a^2+b^2+2\right)\)
\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right)\ge4\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\) (2)
Mà \(4\left[\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\right]\ge4\left(a+b+c+1\right)^2\) (3)(Áp dụng Bunhicopxki và cái ngoặc vuông)
Từ (2) và (3) ta có đpcm.
Sai thì chịu
Xí quên bài 2 b:v
b) Không mất tính tổng quát, giả sử \(\left(a^2-\frac{1}{4}\right)\left(b^2-\frac{1}{4}\right)\ge0\)
Suy ra \(a^2b^2-\frac{1}{4}a^2-\frac{1}{4}b^2+\frac{1}{16}\ge0\)
\(\Rightarrow a^2b^2+a^2+b^2+1\ge\frac{5}{4}a^2+\frac{5}{4}b^2+\frac{15}{16}\)
Hay \(\left(a^2+1\right)\left(b^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{3}{4}\right)\)
Suy ra \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{1}{4}+\frac{1}{2}\right)\left(\frac{1}{4}+\frac{1}{4}+c^2+\frac{1}{2}\right)\)
\(\ge\frac{5}{4}\left(\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c+\frac{1}{2}\right)^2=\frac{5}{16}\left(a+b+c+1\right)^2\) (Bunhiacopxki) (đpcm)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)
Cách nữa cho bài 2:
2a) Ta có: \(4\left(a^2+1+2\right)\left(1+1+\frac{\left(b+c\right)^2}{2}\right)\ge4\left(a+b+c+1\right)^2\)
Hay \(4\left(a^2+3\right)\left(2+\frac{\left(b+c\right)^2}{2}\right)\ge4\left(a+b+c+1\right)^2=VP\)
Như vậy ta quy bài toán về chứng minh: \(\left(b^2+3\right)\left(c^2+3\right)\ge4\left(2+\frac{\left(b+c\right)^2}{2}\right)\)
\(\Leftrightarrow b^2c^2+b^2+c^2+1\ge4bc\Leftrightarrow\left(bc-1\right)^2+\left(b-c\right)^2\ge0\)(đúng)
Đẳng thức xảy ra khi a = b = c = 1
b) Áp dụng BĐT Bunhiacopxki:\(\left(a^2+\frac{1}{4}+\frac{1}{4}+\frac{1}{2}\right)\left(\frac{1}{4}+b^2+c^2+\frac{1}{2}\right)\ge\frac{1}{4}\left(a+b+c+1\right)^2\)
\(\Rightarrow\frac{5}{4}\left(a^2+1\right)\left(b^2+c^2+\frac{3}{4}\right)\ge\frac{5}{16}\left(a+b+c+1\right)^2\)
Từ đó ta có thể quy bài toán về chứng minh: \(\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{4}\left(b^2+c^2+\frac{3}{4}\right)\)
...
Bài 3:Sửa đề a, b, c >0
Có: \(\frac{a^3}{b^2}+\frac{a^3}{b^2}+b\ge3\sqrt[3]{\frac{a^6}{b^3}}=\frac{3a^2}{b}\)
Tương tự: \(\frac{2b^3}{c^2}+c\ge\frac{3b^2}{c};\frac{2c^3}{a^2}+a\ge\frac{3c^2}{a}\)
Cộng theo vế 3 BĐT trên: \(2\left(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\right)+a+b+c\ge3\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\)
\(=2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\)
\(\ge2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+a+b+c\)
Từ đó ta có đpcm.
Cho a, b, c > 0 và abc = 1. CMR:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{3}{4}\)
Giúp mình với các bạn ơiii
Theo bất đẳng thức AM - GM, ta có: \(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge3\sqrt[3]{\frac{a^3}{\left(1+b\right)\left(1+c\right)}.\frac{1+b}{8}.\frac{1+c}{8}}=\frac{3}{4}a\Rightarrow\frac{a^3}{\left(1+b\right)\left(1+c\right)}\ge\frac{3a}{4}-\frac{b+c}{8}-\frac{1}{4}\)Tương tự, ta được: \(\frac{b^3}{\left(1+c\right)\left(1+a\right)}\ge\frac{3b}{4}-\frac{c+a}{8}-\frac{1}{4}\); \(\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{3c}{4}-\frac{a+b}{8}-\frac{1}{4}\)
Cộng vế theo vế ba bất đẳng thức trên, ta được: \(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{4}\left(a+b+c\right)-\frac{3}{4}=\frac{1}{2}\left(a+b+c\right)-\frac{3}{4}\ge\frac{1}{2}.3\sqrt[3]{abc}-\frac{3}{4}=\frac{3}{4}\)Đẳng thức xảy ra khi a = b = c = 1
Cho a, b, c là các số dương biết abc = 1. Chứng minh rằng:
\(\frac{a^3}{\left(b+1\right)\left(c+2\right)}+\frac{b^3}{\left(c+1\right)\left(a+2\right)}+\frac{c^3}{\left(a+1\right)\left(b+2\right)}\ge\frac{1}{2}\)
Ta có \(a+b+c\ge3\sqrt[3]{abc}=3\)
Áp dụng bđt cosi ta có:
\(\frac{a^3}{\left(b+1\right)\left(c+2\right)}+\frac{b+1}{12}+\frac{c+2}{18}\ge3\sqrt[3]{\frac{a^3}{12.18}}=\frac{a}{2}\)
Làm tương tự
=>\(VT+\left(\frac{a+1}{12}+\frac{a+2}{18}\right)+\left(\frac{b+1}{12}+\frac{b+2}{18}\right)+\left(\frac{c+1}{12}+\frac{c+2}{18}\right)\ge\frac{a+b+c}{2}\)
=> \(VT\ge\frac{13}{36}.\left(a+b+c\right)-\frac{7}{12}\ge\frac{13}{36}.3-\frac{7}{12}=\frac{1}{2}\)(ĐPCM)
cho a, b, c dương. chứng minh
\(\frac{1}{a\left(a+b\right)}+\frac{1}{b\left(b+c\right)}+\frac{1}{c\left(c+a\right)}\ge\frac{3}{\sqrt[3]{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
Cho a,b,c>0 và abc=1. CMR:
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)
Chứng minh bất đẳng thức \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
Có: \(\left[\left(\frac{a}{\sqrt{x}}\right)^2+\left(\frac{b}{\sqrt{y}}\right)^2+\left(\frac{c}{\sqrt{z}}\right)^2\right]\left(\sqrt{x}^2+\sqrt{y}^2+\sqrt{z}^2\right)\ge\left(a+b+c\right)^2\) (Bunyakovsky)
\(\Leftrightarrow\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
abc = 1 => a^2.b^2.c^2 = 1
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{a^2b^2c^2}{a^3\left(b+c\right)}+\frac{a^2b^2c^2}{b^3\left(c+a\right)}+\frac{a^2b^2c^2}{c^3\left(a+b\right)}\)
\(=\frac{\left(bc\right)^2}{ab+ac}+\frac{\left(ac\right)^2}{bc+ba}+\frac{\left(ab\right)^2}{ca+cb}\ge\frac{\left(ab+ac+bc\right)^2}{2\left(ab+ac+bc\right)}=\frac{\left(ab+ac+bc\right)}{2}\)
\(\ge\frac{3\sqrt[3]{ab.ac.bc}}{2}\)(Cauchy) \(=\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}a=b=c\\\frac{bc}{ab+ac}=\frac{ac}{bc+ba}+\frac{ab}{ca+cb}\Leftrightarrow\end{cases}a=b=c}\)
Mà abc=1 <=> a^3 = 1 <=> a=1 => b=c=a=1
https://diendantoanhoc.net/topic/80159-ch%E1%BB%A9ng-minh-frac1a2b3cfrac12a3bcfrac13bb2c-leqslant-frac316/
bạn tham khảo ở đây nhé
cách này ổn nè :
với a,b,c \(\in\)R và x,y,z > 0 . Áp dụng BĐT : \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
Dấu " = " xảy ra \(\Leftrightarrow\)\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Ta có : \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{\frac{1}{a^2}}{ab+bc}+\frac{\frac{1}{b^2}}{bc+ab}+\frac{\frac{1}{c^2}}{ac+bc}\)
\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ac\right)}=\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\)( vì abc = 1 )
Mà \(\left(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\right)^3\ge\frac{1}{abc}\)hay \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)
\(\Rightarrow\)\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c = 1
cho a,b,c >0; abc=1.chứng minh
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)
Đặt \(\left(a;b;c\right)=\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\Rightarrow xyz=1\)
\(P=\frac{x^3yz}{y+z}+\frac{xy^3z}{x+z}+\frac{xyz^3}{x+y}=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
\(P\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)
Chứng minh rằng \(\frac{1}{2\sqrt[3]{abc}}+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\forall a,b,c>0\)
\(VT=\frac{\left(\sqrt[3]{abc}\right)^2}{2abc}+\Sigma\frac{a^2}{a^2\left(b+c\right)}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\Sigma a^2\left(b+c\right)+2abc}=\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Cho 3 số thực dương a,b,c thỏa mãn abc=1. Chứng minh rằng:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{3}{4}\)
Bạn tham khảo tại đây:
Câu hỏi của Trần Hữu Ngọc Minh - Toán lớp 9 - Học toán với OnlineMath
Áp dụng BĐT Cosi ta được:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge3\sqrt{\frac{a^3\left(1+b\right)\left(1+c\right)}{\left(1+b\right)\left(1+c\right)64}}=\frac{3a}{4}̸\)
Tương tự \(\hept{\begin{cases}\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{1+a}{8}+\frac{1+c}{8}\ge\frac{3b}{4}\\\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{1+a}{8}+\frac{1+b}{8}\ge\frac{3c}{4}\end{cases}}\)
Cộng theo từng vế BĐT trên ta có:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{3}{4}\ge\frac{a+b+c}{2}\)
Vì \(a+b+c\ge3\sqrt[3]{abc}=3\)do đó:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{3}{4}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{3}{4}\left(đpcm\right)\)
Đẳng thức xảy ra <=> a=b=c