\(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3=\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)\)

\(=a^3+b^3+c^3+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+bc+c^2\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(=a^3+b^3+c^3+3\left(abc+c^2a+b^2c+bc^2+a^2b+ca^2+ab^2+abc\right)\)

\(=a^3+b^3+c^3+3\left[ab\left(a+b+c\right)+bc\left(a+b+c\right)+ca\left(a+b+c\right)\right]\)

\(=a^3+b^3+c^3+3\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\Rightarrow\)\(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b+c\right)\left(ab+bc+ca\right)\)

Lại có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=\left(a-b+b-c+c-a\right)^2\)

\(-2\left[\left(a-b\right)\left(b-c\right)+\left(b-c\right)\left(c-a\right)+\left(c-a\right)\left(a-b\right)\right]\)

\(=-2\left(ab-ca-b^2+bc+bc-ab-c^2+ca+ca-bc-a^2+ab\right)\)

\(=2\left(a^2+b^2+c^2-ab-bc-ca\right)=2\left(a+b+c\right)^2-6\left(ab+bc+ca\right)\)

\(\Rightarrow\)\(P=\frac{\left(a+b+c\right)^3-3\left(a+b+c\right)\left(ab+bc+ca\right)}{2\left(a+b+c\right)^2-6\left(ab+bc+ca\right)}\)

\(=\frac{\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right]}{2\left[\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right]}=\frac{a+b+c}{2}\)

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

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

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

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

\(b,P=\frac{2}{a\left(\sqrt{a}-1\right)}=\frac{2}{25\left(5-1\right)}=\frac{2}{25.4}=\frac{1}{50}\)

Vậy \(P=\frac{1}{50}\)tại \(a=25\)

Với mọi a nguyên dương , 

Ta có: 

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

Biểu thức:

\(B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}\)

\(>2\sqrt{2}-2\sqrt{1}+2\sqrt{3}-2\sqrt{2}+2\sqrt{4}-2\sqrt{3}+...+2\sqrt{25}-2\sqrt{24}\)

\(=-2\sqrt{1}+2\sqrt{25}=-2+10=8\)

Vậy B>8

Ta có:

\(a^3+b^3+c^3\)

\(=\left(\frac{1}{3}a^3+\frac{1}{3}a^3+\frac{1}{3}b^3\right)+\left(\frac{1}{3}b^3+\frac{1}{3}b^3+\frac{1}{3}c^3\right)+\left(\frac{1}{3}c^3+\frac{1}{3}c^3+\frac{1}{3}a^3\right)\)

Áp dụng bất đẳng thức Cô-si cho 3 số không âm ta có:

\(\frac{1}{3}a^3+\frac{1}{3}a^3+\frac{1}{3}b^3\ge3\sqrt[3]{\frac{a^3}{3}\frac{a^3}{3}\frac{b^3}{3}}=\frac{3a^2b}{3}=a^2b\)

Tương tự:

\(\frac{1}{3}b^3+\frac{1}{3}b^3+\frac{1}{3}c^3\ge b^2c\)

\(\frac{1}{3}c^3+\frac{1}{3}c^3+\frac{1}{3}a^3\ge c^2a\)

Cộng vế theo vế ta đc:

\(a^3+b^3+c^3\ge a^2b+b^2c+c^2a\)

Dấu "=" xảy ra khi và chỉ khi a=b=c

Để ý thì thấy đa thức hoán vị: Vì nếu đặt \(f\left(a;b;c\right)=VT-VP\) thì \(f\left(a;b;c\right)=f\left(b;c;a\right)=f\left(c;a;b\right)\) vì vậy ta có thể giả sử \(a=max\left\{a,b,c\right\}\)

\(VT-VP=c\left(\Sigma_{cyc}a^2-\Sigma_{cyc}ab\right)+a\left(a-b\right)\left(a-c\right)+b\left(b-c\right)^2\ge0\)

Đẳng thức xảy ra khi \(a=b=c\)

\(\sqrt{7+2\sqrt{12}}-\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{7+4\sqrt{3}}-\sqrt{5-2\sqrt{6}}\)

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

\(=|2+\sqrt{3}|-|\sqrt{2}-\sqrt{3}|\)

\(=2+\sqrt{3}-\sqrt{3}+\sqrt{2}\)

\(=2+\sqrt{2}\)

\(\sqrt{10-2\sqrt{21}}+\sqrt{9-2\sqrt{14}}\)

\(=\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}+\sqrt{\left(\sqrt{2}-\sqrt{7}\right)^2}\)

\(=|\sqrt{3}-\sqrt{7}|+|\sqrt{2}-\sqrt{7}|\)

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}-\sqrt{2}\)

\(=2\sqrt{7}-\sqrt{3}-\sqrt{2}\)

Với mọi số tự nhiên a> 1 ta có:

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

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

Áp dụng vào bài tập trên ta có:

\(S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{144}}\)

\(>2\sqrt{2}-2\sqrt{1}+2\sqrt{3}-2\sqrt{2}+2\sqrt{4}-2\sqrt{3}+...+2\sqrt{145}-2\sqrt{144}\)

\(=-2\sqrt{1}+2\sqrt{145}>2\left(\sqrt{145}-1\right)>2\left(\sqrt{144}-1\right)=22\)

=> S>22

\(S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{144}}\)

\(< 1+2\sqrt{2}-2\sqrt{1}+2\sqrt{3}-2\sqrt{2}+...+2\sqrt{144}-2\sqrt{143}\)

\(=1-2\sqrt{1}+2\sqrt{144}=23\)

=> S<23

Vậy 22<S<23