Ta có : \(b=\dfrac{c+a}{2}\Rightarrow2b=c+a\Rightarrow a-b=b-c\)

Dó đó : \(P=\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}\right)\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}-\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}-\sqrt{c}\right)}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{a-b}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{b-c}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\) Vì  \(\left(a-b=b-c\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}+\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\dfrac{\sqrt{a}-\sqrt{c}}{b-c}\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\dfrac{a-c}{a-b}=\dfrac{a-c}{a-\dfrac{a+c}{2}}=\dfrac{a-c}{\dfrac{2a-a-c}{2}}=\dfrac{a-c}{\dfrac{a-c}{2}}=2\)

Lời giải:

Áp dụng BĐT Cô-si cho các số dương:

\((a+b)^2+\frac{a+b}{2}=(a+b)[(a+b)+\frac{1}{2}]\)

\(=(a+b)[(a+\frac{1}{4})+(b+\frac{1}{4})]\geq 2\sqrt{ab}(\sqrt{a}+\sqrt{b})=2a\sqrt{b}+2b\sqrt{a}\)

Ta có đpcm

Dấu "=" xảy ra khi $a=b=\frac{1}{4}$

\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)

\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)

\(\sqrt{ab}+\sqrt{4b.c}+2\left(a+c\right)\le\dfrac{1}{2}\left(a+b\right)+\dfrac{1}{2}\left(4b+c\right)+2\left(a+c\right)=\dfrac{5}{2}\left(a+b+c\right)\)

\(\Rightarrow P\ge\dfrac{2}{5}\left(\dfrac{1}{a+b+c}-\dfrac{1}{\sqrt{a+b+c}}\right)=\dfrac{2}{5}\left(\dfrac{1}{\sqrt{a+b+c}}-\dfrac{1}{2}\right)^2-\dfrac{1}{10}\ge-\dfrac{1}{10}\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}a+b+c=4\\a=b=\dfrac{c}{4}\end{matrix}\right.\) em tự giải ra a;b;c

\(P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)

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

\(=x-\sqrt{x}+1\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^3+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=4\end{matrix}\right.\) \(\Rightarrow a+b=7\)

`sqrta+sqrtb+sqrtc=2`

`<=>(sqrta+sqrtb+sqrtc)^2=4`

`<=>a+b+c+2sqrt{ab}+2sqrt{bc}+2sqrt{ca}=4`

`<=>2sqrt{ab}+2sqrt{bc}+2sqrt{ca}=4-(a+b+c)=4-2-2`

`<=>sqrt{ab}+sqrt{bc}+sqrt{ca}=1`

`=>a+1=a+sqrt{ab}+sqrt{bc}+sqrt{ca}=sqrta(sqrta+sqrtb)+sqrtc(sqrta+sqrtb)=(sqrta+sqrtb)(sqrta+sqrtc)`

Tương tự:`b+1=(sqrtb+sqrta)(sqrtb+sqrtc)`

`c+1=(sqrtc+sqrta)(sqrtc+sqrtb)`

`=>VT=sqrta/((sqrta+sqrtb)(sqrta+sqrtc))+sqrtb/((sqrtb+sqrta)(sqrtb+sqrtc))+sqrtc/((sqrtc+sqrta)(sqrtc+sqrtb))`

`=>VT=(sqrta(sqrtb+sqrtc)+sqrtb(sqrtc+sqrta)+sqrtc(sqrta+sqrtb))/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=(sqrt{ab}+sqrt{ac}+sqrt{bc}+sqrt{ab}+sqrt{ac}+sqrt{bc})/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=(2(sqrt{ab}+sqrt{bc}+sqrt{ca}))/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=2/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`

`=2/\sqrt{[(sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta)]^2}`

`=2/\sqrt{(sqrta+sqrtb)(sqrta+sqrtc)(sqrtb+sqrta)(sqrtb+sqrtc)(sqrtc+sqrta)(sqrtc+sqrtb)}`

`=2/\sqrt{(1+a)(1+b)(1+c)}=>đpcm`

ráng chờ thầy nguyễn việt lâm  onl r nhờ nghen:>

\(\sqrt{\left(16+9\right)\left(16a^2b^2+9\right)}\ge\sqrt{\left(16ab+9\right)^2}=16ab+9\)

\(\Rightarrow\sqrt{16a^2b^2+9}\ge\dfrac{1}{5}\left(16ab+9\right)\)

\(\Rightarrow P\ge\dfrac{1}{5}\left(16ab+9\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{1}{5}\left[16\left(a+b\right)+9\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right]\)

\(P\ge\dfrac{1}{5}\left[32+9.\dfrac{4}{a+b}\right]=\dfrac{1}{5}\left[32+\dfrac{9.4}{2}\right]=10\)

\(P_{min}=10\) khi \(a=b=1\)

a: \(=\dfrac{\left(2+\sqrt{3}-1\right)\cdot\sqrt{3}}{\sqrt{7+4\sqrt{3}-2-\sqrt{3}+1}}\)

\(=\dfrac{\left(\sqrt{3}+1\right)\cdot\sqrt{3}}{\sqrt{6+3\sqrt{3}}}=\left(\sqrt{3}+1\right)\cdot\sqrt{\dfrac{1}{2\sqrt{3}+3}}\)

\(=\left(\sqrt{3}+1\right)\cdot\sqrt{\dfrac{\sqrt{3}\left(2-\sqrt{3}\right)}{3}}\)

\(=\left(\sqrt{3}+1\right)\cdot\sqrt{\dfrac{2-\sqrt{3}}{\sqrt{3}}}\)

\(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)\left(4+2\sqrt{3}\right)}{\sqrt{3}}}\)

\(=\sqrt{\dfrac{8-6}{\sqrt{3}}}=\sqrt{\dfrac{2\sqrt{3}}{3}}\)

c: \(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}+...-\sqrt{1994}+\sqrt{1995}\)

\(=\sqrt{1995}-1\)