Why do you say maintenance page ? 

 We have : \(\sqrt{xy\left(x-y\right)}=x+y\Leftrightarrow xy\left(x-y\right)^2=\left(x+y\right)^2\)

\(xy\left(x-y\right)^2=\dfrac{1}{4}.4xy\left[\left(x+y\right)^2-4xy\right]\le\dfrac{\left(x+y\right)^4}{16}\)

so \(\left(x+y\right)^4\ge16\left(x+y\right)^2\)  \(\Leftrightarrow p^4-16p^2\ge0\Leftrightarrow p\ge4\)

Equal sign occurs \(\Leftrightarrow x=2+\sqrt{2};b=2-\sqrt{2}\)

MathYou not Englishyou 

We have : \(x^2-180x+8102=\left(x^2-180x+8100\right)+2=\left(x-90\right)^2+2\ge2\forall x\left(1\right)\)

Applying the Bunhiacopxki inequality , we have : 

\(\left(\sqrt{x-89}+\sqrt{91-x}\right)^2\le\left(1+1\right)\left(x-89+91-x\right)\)

\(\Rightarrow\left(\sqrt{x-89}+\sqrt{91-x}\right)^2\le2.2=4\)

\(\Rightarrow\sqrt{x-89}+\sqrt{91-x}\le2\left(2\right)\)

Because \(\sqrt{x-89}+\sqrt{91-x}=x^2-180x+8102\)

So ( 1 ) ; ( 2 ) ; we have : \(\sqrt{x-89}+\sqrt{91-x}=x^2-180x+8102=2\)

Equal sign occurs \(\Leftrightarrow\sqrt{x-89}=\sqrt{91-x};x-90=0\left(3\right)\)

We have : \(\sqrt{x-89}=\sqrt{91-x}\)  \(\Leftrightarrow x-89=91-x\Leftrightarrow x=90\left(4\right)\)

 ( 3 ) ; ( 4 ) \(\Rightarrow x=90\) is the result of the equation

Tran Anh ??? Le Quoc Tran Anh ??? 

Put \(A=\dfrac{ab}{a+b-c}+\dfrac{bc}{b+c-a}+\dfrac{ac}{c+a-b}\)

Because a ; b ; c are the length of a triangle so \(a+b-c;b+c-a;c+a-b>0\)

Put \(a+b-c=x;b+c-a=y;c+a-b=z\)

\(\Rightarrow\dfrac{x+y}{2}=b;\dfrac{y+z}{2}=c;\dfrac{x+z}{2}=a;a+b+c=x+y+z\) 

We have : \(\dfrac{\left(x+y\right)\left(x+z\right)}{2.2x}+\dfrac{\left(x+y\right)\left(y+z\right)}{2.2y}+\dfrac{\left(x+z\right)\left(y+z\right)}{2.2z}\)

\(=\dfrac{x\left(x+y+z\right)+yz}{4x}+\dfrac{y\left(x+y+z\right)+xz}{4y}+\dfrac{z\left(x+y+z\right)+xy}{4z}\)

\(=\dfrac{x+y+z}{4}+\dfrac{x+y+z}{4}+\dfrac{x+y+z}{4}+\dfrac{yz}{4x}+\dfrac{xz}{4y}+\dfrac{xy}{4z}\)

\(=\dfrac{3\left(x+y+z\right)}{4}+\dfrac{y^2z^2}{4xyz}+\dfrac{x^2z^2}{4xyz}+\dfrac{x^2y^2}{4xyz}\)

Applying the inequality \(a^2+b^2+c^2\ge ab+bc+ac\) , we have :

\(A\ge\dfrac{3\left(x+y+z\right)}{4}+\dfrac{xyz\left(x+y+z\right)}{4xyz}=x+y+z=a+b+c\)

Equal sign occurs \(\Leftrightarrow x=y=z\Leftrightarrow a=b=c\)

I don't know why 

\(\dfrac{5}{x}+\dfrac{5}{y}\ge5.\dfrac{4}{x+y}\ge5.\dfrac{4}{10}\)

Because I think \(\dfrac{4}{x+y}\le\dfrac{4}{10}=\dfrac{2}{5}\)

Conandtb spam 

You can ask your teacher 

??? Hack ??? 

My answer is :

 \(200cm^2\)

:D

2067 mm + 478 cm - 0,1356 m 

= 2,067 m + 4,78 m - 0 , 1356 m 

= 6 , 7114 m 

= 0 , 0067114 km 

My answer is : D  3 , 3535