Áp dụng BĐT Bunhiacopski ta dễ có:

\(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge3\sqrt{3\left(2a+2b+2c\right)}=3\sqrt{6\left(a+b+c\right)}\)

\(=3\sqrt{24}>4\)

Đẳng thức xảy ra tại a=b=c=⁴/₃

a.  \(x^2\left(y-1\right)+y^2\left(x-1\right)=1\)

<=> \(x^2y+y^2x-\left(x^2+y^2\right)=1\)

<=> \(xy\left(x+y\right)-\left(x+y\right)^2+2xy=1\)

Đặt: x + y = u; xy = v => u; v là số nguyên

Ta có: uv - \(u^2+2v=1\)

<=> \(u^2-uv-2v+1=0\) 

<=> \(u^2+1=v\left(2+u\right)\)

=> \(u^2+1⋮2+u\)

=> \(u^2-4+5⋮2+u\)

=> \(5⋮2-u\)

=> 2 - u = 5; 2 - u = -5; 2- u = 1; 2- u = -1 

Mỗi trường hợp sẽ tìm đc v 

=> x; y 

\(E=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{\left(4+\sqrt{15}\right)^2}.\sqrt{\left(\sqrt{10}-\sqrt{6}\right)^2}.\frac{4^2-15}{\sqrt{4+\sqrt{15}}}\)

\(=\sqrt{4+\sqrt{15}}.\sqrt{10+6-2\sqrt{10}.\sqrt{6}}\)

\(=\sqrt{4+\sqrt{15}}.\sqrt{16-2\sqrt{60}}\)

\(=\sqrt{4+\sqrt{15}}.\sqrt{4\left(4-\sqrt{15}\right)}\)

\(=2\sqrt{\left(4+\sqrt{15}\right).\left(4-\sqrt{15}\right)}\)

\(=2\sqrt{16-15}=2\)

\(D=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(\Rightarrow D^2=4+\sqrt{10+2\sqrt{5}}+2\sqrt{4+\sqrt{10+2\sqrt{5}}}.\sqrt{4-\sqrt{10+2\sqrt{5}}}+4-\sqrt{10+2\sqrt{5}}\)

\(=8+2\sqrt{4^2-\left(\sqrt{10+2\sqrt{5}}\right)^2}\)

\(=8+2\sqrt{16-10-2\sqrt{5}}\)

\(=8+2\sqrt{6-2\sqrt{5}}\)

\(=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)

\(\Rightarrow D=\sqrt{5}+1\)

\(C=\sqrt{7-3\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

\(\Rightarrow C.\sqrt{2}=\sqrt{14-6\sqrt{5}}+\sqrt{6-\sqrt{5}}\)

\(=\sqrt{9-2.3.\sqrt{5}+5}+\sqrt{5-2.\sqrt{5}.1+1}\)

\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=3-\sqrt{5}+\sqrt{5}-1=2\)

\(\Rightarrow C=\sqrt{2}\)

\(B=\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}< 0\)

\(\Rightarrow B^2=4-\sqrt{15}-2\sqrt{4-\sqrt{15}}.\sqrt{4+\sqrt{15}}+4+\sqrt{15}\)

\(=8-2\sqrt{4^2-\left(\sqrt{15}\right)^2}=8-2=6\)

\(\Rightarrow B=-\sqrt{6}\)

(Vì \(\sqrt{4-\sqrt{15}}< \sqrt{4+\sqrt{15}}\)nên B nhận dấu âm)

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

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

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

Cho mình bổ sung nha, nãy bấm nhầm gửi lun

Xét \(\sqrt{2}< 2\Rightarrow2+\sqrt{2}< 4\Rightarrow\sqrt{2+\sqrt{2}}< 2\Rightarrow2+\sqrt{2+\sqrt{2}}< 4\)

\(\Rightarrow\sqrt{2+\sqrt{2+\sqrt{2}}}< 2\Rightarrow2+\sqrt{2+\sqrt{2+\sqrt{2}}}< 4\)

\(\Rightarrow\frac{1}{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}>\frac{1}{4}\)

\(\Rightarrow B>\frac{1}{4}\)