Maximum

\(x^2+5y^2-4xy-x+2y-6=0\)

\(\Leftrightarrow x^2-4xy+4y^2-\left(x-2y\right)-6=-y^2\)

\(\Leftrightarrow\left(x-2y\right)^2-\left(x-2y\right)-6=-y^2\le0\)

\(\Leftrightarrow\left(x-2y+2\right)\left(x-2y-3\right)\le0\)

\(\Leftrightarrow-2\le x-2y\le3\)

Use Cauchy's inequality for positive numbes a,b,c.

\(a+b\ge2\sqrt{ab},b+c\ge2\sqrt{bc},c+a\ge2\sqrt{ca}\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{ab}\cdot\sqrt{bc}\cdot\sqrt{ca}=8abc\)

\(\Rightarrow A=\dfrac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{1}{8}\)

\(MaxA=\dfrac{1}{8}\Leftrightarrow a=b=c>0\)

Use Cauchy's inequality for positive numbes a,b,c.

a+b≥2ab−−√,b+c≥2bc−−√,c+a≥2ca−−√

⇒(a+b)(b+c)(c+a)≥8ab−−√⋅bc−−√⋅ca−−√=8abc

⇒A=abc(a+b)(b+c)(c+a)≤18

MaxA=18⇔a=b=c>0

We have :

\(x^2-2x+5=x^2-2x+1+4\)

                      \(=\left(x-1\right)^2+4\ge4\)

\(\Rightarrow H=\dfrac{1}{x^2-2x+5}\le\dfrac{1}{4}\)

\(\Rightarrow Max_H=\dfrac{1}{4}\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\)

We have :

x2−2x+5=x2−2x+1+4

                      =(x−1)2+4≥4

⇒H=1x2−2x+5≤14

⇒MaxH=14

⇔(x−1)2=0

⇔x=1

(*) 1+N=.....=x2/(x-2)2+4

We have x2 ≥

 0 , (x-2)2+4 ≥

 4 > 0 

So 1+N ≥

 0 => N ≥

 -1 ;equality : x=0 

(*)1-N=....=(x-4)2/(x-2)2+4 

....... -> N ≤

 1 , equality : x=4

There is something is wrong here 

Change to :

We have a2+b2≥5

=> 9−(a2+b2)≤4

=> 2ab≤4

=> ab≤2

<=> a2b2≥4

=> 4a2b2≥16

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