Ta co :\(M-N=\left(a+b-1\right)-\left(b+c-1\right)\)

            \(M-N=a+b-1-b-c+1\)

            \(M-N=\left(a-c\right)+\left(b-b\right)+\left(1-1\right)\)
            \(M-N=a-c+0+0\)

            \(M-N=a-c\)

Vi M>N \(\Rightarrow\) M-N la so nguyen duong \(\Rightarrow\) hieu a-c la so nguyen duong

TICK CHO MINH NHA .MINH LA NGUOI GIAI NHANH NHAT DO VA RAT CHI TIET TICK NHA BAN!!!!!!

5 so nguyen do deu bang am 3

tất cả 5 số nguyên đó đêuf bằng -3

Lời giải:

Vì tam giác có chu vi bằng $1$ nên $a+b+c=1$

\(\Rightarrow 1-a, 1-b, 1-c>0\)

Thay vào biểu thức đã cho:

\(\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=\frac{3}{2}\)

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\right)[a(1-a)+b(1-b)+c(1-c)]\geq (a+b+c)^2\)

\(\Leftrightarrow \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq \frac{(a+b+c)^2}{(a+b+c)-(a^2+b^2+c^2)}\)

\(\Leftrightarrow \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq \frac{1}{1-(a^2+b^2+c^2)}\)

Áp dụng BĐT Bunhiacopxky: \((a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2=1\)

\(\Leftrightarrow a^2+b^2+c^2\geq \frac{1}{3}\)

\(\Rightarrow 1-(a^2+b^2+c^2)\leq \frac{2}{3}\)

Suy ra \( \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq \frac{1}{1-(a^2+b^2+c^2)}\geq \frac{1}{\frac{2}{3}}=\frac{3}{2}\)

Dấu bằng xảy ra khi \(\frac{a}{1}=\frac{b}{1}=\frac{c}{1}\Leftrightarrow a=b=c\) hay tam giác $ABC$ đều.

Cách khác:v

Giải: \(gt:\left\{{}\begin{matrix}a;b;c>0\\a+b+c=1\end{matrix}\right.\)

\(pt\Leftrightarrow\dfrac{a}{\left(a+b+c\right)-a}+\dfrac{b}{\left(a+b+c\right)-b}+\dfrac{c}{\left(a+b+c\right)-c}=\dfrac{3}{2}\)

\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{3}{2}\)

\(Nesbit:\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)

Dấu "=" \(a=b=c\Leftrightarrow\Delta ABC\) đều

a=b=c=2 thay vào ra min cái này là tay tui tự gõ ra a=b=c=2 chả có bước nào. còn chi tiết sau nhớ nhắc tui làm :D

Áp dụng BĐT Mincopxki và AM-GM có:

\(T=\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{a^2}}\)

\(\ge\sqrt{\left(a+b+c\right)^2+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)

\(\ge\sqrt{\left(a+b+c\right)^2+\frac{81}{\left(a+b+c\right)^2}}\)

\(=\sqrt{\frac{81}{\left(a+b+c\right)^2}+\frac{\left(a+b+c\right)^2}{16}+\frac{15\left(a+b+c\right)^2}{16}}\)

\(=\sqrt{2\sqrt{\frac{81}{\left(a+b+c\right)^2}\cdot\frac{\left(a+b+c\right)^2}{16}}+\frac{15\cdot6^2}{16}}\)

\(=\sqrt{2\sqrt{\frac{81}{16}}+\frac{15\cdot6^2}{16}}=\frac{3\sqrt{17}}{2}\)

Khi \(a=b=c=2\)

A=\(\frac{1}{2}.\left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+.......+\frac{1}{9240}\right)>\frac{57}{462}\)ta nhóm 6 là ngoài

A=\(\frac{1}{2}.6.\left(1+\frac{1}{4}+\frac{1}{10}+.........+\frac{1}{9240}\right)>\frac{57}{462}\)

A=3.(1+\(\frac{1}{1540}\))

A=3.\(\frac{1541}{1540}\)

A=3 

\(\Rightarrow\)3>\(\frac{1541}{1540}\)

quen 

3>\(\frac{57}{462}\)

\(\dfrac{1}{40}km^2+\dfrac{1}{52}km^2=\dfrac{23}{520}km^2\approx44230.77m^2\)