Có :

\(\left(a^2+4b^2+9c^2\right).\left(1+\frac{1}{4}+\frac{1}{9}\right)\ge\left(a+b+c\right)^2\)

\(\Leftrightarrow\frac{49}{36}\ge\left(a+b+c\right)^2\)

\(\Rightarrow A\le\frac{7}{6}\)

c2 : \(\frac{36a^2}{36}+\frac{36b^2}{9}+\frac{36c^2}{4}\ge\frac{\left(6a+6b+6c\right)^2}{49}=\frac{6^2\left(a+b+c\right)^2}{7^2}\)

\(< =>\frac{6^2\left(a+b+c\right)^2}{7^2}\le1< =>a+b+c\le\frac{7}{6}\)

ta có 

\(4\left(x^2+xy+y^2\right)\ge3\left(x+y\right)^2\Leftrightarrow\left(x-y\right)^2\ge0\) vì thế \(\sqrt{x^2+xy+y^2}\ge\frac{\sqrt{3}}{2}\left(x+y\right)\)

hoàn toàn tương tự ta sẽ có 

\(P\ge\frac{\sqrt{3}}{2}\left(x+y\right)+\frac{\sqrt{3}}{2}\left(y+z\right)+\frac{\sqrt{3}}{2}\left(x+z\right)\)

hay

\(P\ge\sqrt{3}\left(x+y+z\right)=3\sqrt{3}\)

dấu bằng xảy ra khi x=y=z=1

 Ta có: \(P=\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\)\(=\sqrt{\frac{3}{4}\left(x+y\right)^2+\frac{1}{4}\left(x-y\right)^2}+\sqrt{\frac{3}{4}\left(y+z\right)^2+\frac{1}{4}\left(y-z\right)^2}+\sqrt{\frac{3}{4}\left(z+x\right)^2+\frac{1}{4}\left(x-y\right)^2}\)\(\ge\sqrt{\frac{3}{4}\left(x+y\right)^2}+\sqrt{\frac{3}{4}\left(y+z\right)^2}+\sqrt{\frac{3}{4}\left(z+x\right)^2}=\sqrt{3}\left(x+y+z\right)=3\sqrt{3}\)

Đẳng thức xảy ra khi x = y = z = 1

\(E=\left(\frac{\sqrt{x}}{x\sqrt{x}-1}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}+\frac{x\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x\sqrt{x}-1\right)}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

\(=\frac{x-\sqrt{x}+x\sqrt{x}-1}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)

\(=\frac{-\sqrt{x}\left(1-x\right)+\left(x-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)

Dể rồi làm nốt nhé, ngại v 

bỏ chữ x đầu nhá mình ghi nhầm :>

\(P=a+\frac{1}{a}=\frac{a}{4}+\frac{1}{a}+\frac{3a}{4}\ge2\sqrt{\frac{a}{4}\cdot\frac{1}{a}}+\frac{3a}{4}\ge2\sqrt{\frac{1}{4}}+\frac{3\cdot2}{4}=\frac{5}{2}\)

Đẳng thức xảy ra <=> a = 2

=> MinP = 5/2, đạt được khi a = 2

Xét hiệu : \(P-\left(2+\frac{1}{2}\right)=a+\frac{1}{a}-2-\frac{1}{2}\)

                                               \(=\left(a-2\right)+\left(\frac{1}{a}-\frac{1}{2}\right)\)

                                               \(=\left(a-2\right)+\frac{2-a}{2a}\)

                                              \(=\left(a-2\right)\left(1-\frac{1}{2a}\right)\)

Vì \(a\ge2\) \(\Rightarrow\hept{\begin{cases}a-2\ge0\\1-\frac{1}{2a}>0\end{cases}}\)

\(\Rightarrow\left(a-2\right)\left(1-\frac{1}{2a}\right)\ge0\)

\(\Rightarrow P\ge2+\frac{1}{2}\)

Dẫu "=" xảy ra \(\Leftrightarrow a=2\)

Vậy \(MinP=2+\frac{1}{2}=\frac{5}{2}\Leftrightarrow a=2\)

\(P=\frac{1}{\sqrt{x}+1}+\frac{10}{2\sqrt{x}+1}-\frac{5}{2x+3\sqrt{x}+1}\)

\(=\frac{1}{\sqrt{x}+1}+\frac{10}{2\sqrt{x}+1}-\frac{5}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{2\sqrt{x}+1+10\left(\sqrt{x}+1\right)-5}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{2\sqrt{x}+1+10\sqrt{x}+10-5}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{6}{\sqrt{x}+1}\)

b) Để P nguyên tố thì  \(\frac{6}{\sqrt{x}+1}\) nguyên tố 

Để \(P\inℕ^∗\) thì  \(\sqrt{x}+1\inƯ\left(6\right)\) 

Mà P nguyên tố \(\Rightarrow\frac{6}{\sqrt{x}+1}=\left\{2;3\right\}\Rightarrow\sqrt{x}+1=\left\{2;3\right\}\)

Với \(\sqrt{x}+1=2\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)

Với \(\sqrt{x}+1=3\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

Vậy ...........