\(0< a;b;c< 1\Rightarrow\left\{{}\begin{matrix}a^3< a^2< a< 1\\b^3< b^2< b< 1\\c^3< c^2< c< 1\end{matrix}\right.\)
\(\Rightarrow\left(1-a\right)\left(1-b^2\right)>0\Rightarrow1+ab^2>a+b^2>a^3+b^3\)
Tương tự: \(1+b^2c>b^3+c^3\); \(1+ca^2>a^3+c^3\)
Cộng vế với vế: \(3+a^2b+b^2c+c^2a>2a^3+2b^3+2c^3\)
Đẳng thức không xảy ra
Cho a,b,c>0 và a+b+c=3.Chứng minh
\(a^2b^2+b^2c^2+c^2a^2\le3\)
Cho a,b,c là độ dài 3 cạnh tam giác. Tìm GTNN của
P=\(\sqrt{\dfrac{2a}{2b+2c-a}}+\sqrt{\dfrac{2b}{2c+2a-b}}+\sqrt{\dfrac{2c}{2a+2b-c}}\)
CMR với mọi số nguyên a,b,c ta đều có BĐT:
\(\dfrac{a^2}{\left(2a+b\right)\left(2a+c\right)}+\dfrac{b^2}{\left(2b+a\right)\left(2b+c\right)}+\dfrac{c^2}{\left(2c+a\right)\left(2c+b\right)}\le\dfrac{1}{3}\)
cmr với mọi số thực a, b, c dươngta đều có bđt
\(\dfrac{a^2}{\left(2a+b\right)\left(2a+c\right)}+\dfrac{b^2}{\left(2b+a\right)\left(2b+c\right)}+\dfrac{a^2}{\left(2c+a\right)\left(2c+b\right)}\)<=3
cho a,b,c> 0 . Cmr:
\(2\left(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\right)\ge1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}\)
Cho tam giác ABC có a=3, b=6, c=\(\sqrt[]{17}\)
Cmr : \(\sin^2A+sin^2B=3sin^2C\)
Source of Question: Câu hỏi của Hiếu Cao Huy - Toán lớp 9 | Học trực tuyến
Xét pt (1): \(\Delta=b^2-4ac\)
\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}\); \(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}\)
Xét pt (2) : \(\Delta=b^2-4ac\)
\(y_1=\dfrac{-b+\sqrt{\Delta}}{2c}\) ; \(y_2=\dfrac{-b-\sqrt{\Delta}}{2c}\)
Thay vào M:
\(M=\dfrac{\left(-b+\sqrt{\Delta}\right)^2}{4a^2}+\dfrac{\left(-b-\sqrt{\Delta}\right)^2}{4a^2}+\dfrac{\left(-b+\sqrt{\Delta}\right)^2}{4c^2}+\dfrac{\left(-b-\sqrt{\Delta}\right)^2}{4c^2}\)
\(=\dfrac{b^2-2b\sqrt{\Delta}+\Delta}{4a^2}+\dfrac{b^2+2b\sqrt{\Delta}+\Delta}{4a^2}+\dfrac{b^2-2b\sqrt{\Delta}+\Delta}{4c^2}+\dfrac{b^2+2b\sqrt{\Delta}+\Delta}{4c^2}\)
\(=\dfrac{2b^2+2\Delta}{4a^2}+\dfrac{2b^2+2\Delta}{4c^2}=\dfrac{b^2+\Delta}{2a^2}+\dfrac{b^2+\Delta}{2c^2}=\dfrac{b^2c^2+\Delta c^2}{2a^2c^2}+\dfrac{a^2b^2+\Delta a^2}{2a^2c^2}\)
\(=\dfrac{b^2\left(a^2+c^2\right)+\Delta\left(a^2+c^2\right)}{2a^2c^2}=\dfrac{\left(b^2+\Delta\right)\left(a^2+c^2\right)}{2a^2c^2}=\dfrac{\left(b^2+b^2-4ac\right)\left(a^2+c^2\right)}{2a^2c^2}\)
\(=\dfrac{\left(2b^2-4ac\right)\left(a^2+c^2\right)}{2a^2c^2}=\dfrac{\left(b^2-2ac\right)\left(a^2+c^2\right)}{a^2c^2}=\dfrac{a^2b^2-2a^3c+b^2c^2-2ac^3}{a^2c^2}\)
\(=\dfrac{a^2b^2}{a^2c^2}+\dfrac{b^2c^2}{a^2c^2}-\dfrac{2a^3c}{a^2c^2}-\dfrac{2ac^3}{a^2c^2}=\dfrac{b^2}{c^2}+\dfrac{b^2}{a^2}-\dfrac{2a}{c}-\dfrac{2c}{a}\)
\(=\left(\dfrac{b^2}{c^2}-\dfrac{2ac}{c^2}\right)+\left(\dfrac{b^2}{a^2}-\dfrac{2ac}{a^2}\right)=\dfrac{b^2-2ac}{c^2}+\dfrac{b^2-2ac}{a^2}\)
\(=\left(b^2-2ac\right)\left(\dfrac{1}{c^2}+\dfrac{1}{a^2}\right)\)
Thanks a lots for your answering ^^!
Hiếu Cao Huy: Wait together!
Cho a,b,c >0 thõa mãn a+b+c = 3
\(CMR:\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}\ge1\)
Cho a,b,c là số dương thỏa mãn \(a^2+b^2+c^2=3\) . Chứng minh rằng
a/ \(a^2b+b^2c+c^2a\le3\)
b/ \(\dfrac{ab}{3+c^2}+\dfrac{bc}{3+a^2}+\dfrac{ca}{3+b^2}\le\dfrac{3}{4}\)
