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

Ta có: \(30< 36\)

=> \(\sqrt{30}< \sqrt{36}=6\)

=> \(\sqrt{30+\sqrt{30}}< \sqrt{30+6}=6\)

=> \(\sqrt{30+\sqrt{30+\sqrt{30}}}< \sqrt{30+6}=6\)

Cứ tiếp tực như vậy ta sẽ so sánh đc:

\(\sqrt{30+\sqrt{30+\sqrt{30+...+\sqrt{30}}}}< 6\)

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

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

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

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

Vậy với \(a>0;a\ne1\) thì \(P=\frac{-2}{\sqrt{a}+1}\)

Để \(P=\frac{-1}{4}\Leftrightarrow\frac{-2}{\sqrt{a}+1}=\frac{-1}{4}\)

  \(\Leftrightarrow-\sqrt{a}-1=-8\)

\(\Leftrightarrow-\sqrt{a}=-7\)

\(\Leftrightarrow a=49\)

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

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

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

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

khi P = - 1/4

\(\Leftrightarrow\frac{-2}{1+\sqrt{a}}=\frac{-1}{4}\)

\(\Leftrightarrow1+\sqrt{a}=8\)

\(\Leftrightarrow\sqrt{a}=7\)

\(\Leftrightarrow a=49\)