We have :  primes from 20 to 30 are 23,29

=> total : 23 + 29 = 52

The answer í 52

The volume of the rectangular prism is:

      5 x 7 x 3 = 105(cm²)

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

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 \(\Leftrightarrow x=y=z\Leftrightarrow a=b=c\)

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\(P=\left(\dfrac{30}{x}+\dfrac{6x}{5}\right)+\left(\dfrac{y}{5}+\dfrac{5}{y}\right)+\dfrac{4}{5}\left(x+y\right)\ge2.\sqrt{\dfrac{30}{x}+\dfrac{6x}{5}}+2.\sqrt{\dfrac{y}{5}+\dfrac{5}{y}}+\dfrac{4}{5}.10\)

\(P=2.6+2.1+8=22\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{30}{x}=\dfrac{6x}{5}\\\dfrac{y}{5}=\dfrac{5}{y}\\x+y=10\end{matrix}\right.=>\left\{{}\begin{matrix}x^2=25\\y^2=25\\x+y=10\end{matrix}\right.\)

\(=>x=y=5\)

\(Pmin=22< =>x=y=5\)

\(P=2x+y+\dfrac{30}{x}+\dfrac{5}{y}=\left(x+y\right)\left(\dfrac{5}{x}+\dfrac{5}{y}\right)+\left(\dfrac{25}{x}+x\right)\)

\(\left\{{}\begin{matrix}x>0=>x=\left(\sqrt{x}\right)^2\\y>0=>y=\left(\sqrt{y}\right)^2\end{matrix}\right.\)

\(P1=x+y\ge10\)

\(P2=\dfrac{5}{x}+\dfrac{5}{y}=5.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge5.\dfrac{4}{x+y}=\dfrac{5.4}{10}=2\) khi \(x=y\)

\(P3=\dfrac{25}{x}+x\ge2.\sqrt{\dfrac{25}{x}.x}=2.5=10\) khi \(x=5\)

\(P=\sum P\ge10+2+10=22\) khi \(\left(x;y\right)=\left(5;5\right)\)

Study well

Reference Lê Anh Duy