\(ab\le\frac{a^2+b^2}{2}\le\frac{16}{2}=8\)

Ta có: \(N^2=\left(a\sqrt{9b\left(a+8b\right)}+b\sqrt{9a\left(b+8a\right)}\right)^2\)

\(\le\left(a^2+b^2\right)\left[9b\left(a+8b\right)+9a\left(b+8a\right)\right]\)

\(\le16\left(18ab+72\left(a^2+b^2\right)\right)\le16\left(18.8+72.16\right)\)

\(=20736\)

=> \(N\le144\)

Dấu "=" xảy ra <=> a = b = \(\sqrt{8}\)

Vậy max N = 144 tại a = b = \(\sqrt{8}\)

By C-S and AM-GM's inequality

\(M=a\left(9b\left(a+8b\right)\right)^{\dfrac{1}{2}}+b\left(9a\left(b+8a\right)\right)^{\dfrac{1}{2}}\)

\(\le\left(\left(a^2+b^2\right)\left(9b\left(a+8b\right)+9a\left(b+8a\right)\right)\right)^{\dfrac{1}{2}}\)

\(=\left(\left(a^2+b^2\right)\left(18ab+72b^2+72a^2\right)\right)^{\dfrac{1}{2}}\)

\(=\left(\left(a^2+b^2\right)\left(18\cdot\dfrac{a^2+b^2}{2}+72b^2+72a^2\right)\right)^{\dfrac{1}{2}}\)

\(=\left(16\cdot\left(18\cdot\dfrac{16}{2}+72\cdot16\right)\right)^{\dfrac{1}{2}}=144\)

\("="\Leftrightarrow a=b=2\sqrt{2}\)

Ta có \(2ab\le a^2+b^2\)

Áp dụng BĐT Bunhia:

\(M^2\le\left(a^2+b^2\right)\left(b\left(a+8b\right)+a\left(b+8a\right)\right)\)

\(\Rightarrow M^2\le\left(a^2+b^2\right)\left(2ab+8b^2+8a^2\right)\)

\(\Rightarrow M^2\le\left(a^2+b^2\right)\left(8a^2+8b^2+a^2+b^2\right)\)

\(\Rightarrow M^2\le9\left(a^2+b^2\right)^2\Rightarrow M\le3\left(a^2+b^2\right)=48\)

\(\Rightarrow M_{max}=48\) khi \(a=b=2\sqrt{2}\)

2M\(\le\)a(9b+4a+5b)+b(9a+4b+5a)  (AM-GM)

     =4(a²+b²)+28ab\(\le\)4(a²+b²)+14(a²+b²)  (AM-GM)

                                =36 (do a²+b²=2)

=> M \(\le\)18

 Dấu bằng có <=> a=b=1

Bài 1:

Áp dụng BĐT Bunhiacopxky:

\(M^2=(a\sqrt{9b(a+8b)}+b\sqrt{9a(b+8a)})^2\)

\(\leq (a^2+b^2)(9ab+72b^2+9ab+72a^2)\)

\(\Leftrightarrow M^2\leq (a^2+b^2)(72a^2+72b^2+18ab)\)

Áp dụng BĐT AM-GM: \(a^2+b^2\geq 2ab\Rightarrow 18ab\leq 9(a^2+b^2)\)

Do đó, \(M^2\leq (a^2+b^2)(72a^2+72b^2+9a^2+9b^2)=81(a^2+b^2)^2\)

\(\Leftrightarrow M\leq 9(a^2+b^2)\leq 144\)

Vậy \(M_{\max}=144\Leftrightarrow a=b=\sqrt{8}\)

Bài 6:

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

Vì \(a>1\rightarrow a-1>0\). Do đó áp dụng BĐT Am-Gm cho số dương\(a-1,\frac{1}{a-1}\) ta có:

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

\(\Rightarrow a+\frac{1}{a-1}=1+(a-1)+\frac{1}{a-1}\geq 3\) (đpcm)

Dấu bằng xảy ra khi \(a-1=1\Leftrightarrow a=2\)

Bài 3:

Xét \(\sqrt{a^2+1}\). Vì \(ab+bc+ac=1\) nên:

\(a^2+1=a^2+ab+bc+ac=(a+b)(a+c)\)

\(\Rightarrow \sqrt{a^2+1}=\sqrt{(a+b)(a+c)}\)

Áp dụng BĐT AM-GM có: \(\sqrt{(a+b)(a+c)}\leq \frac{a+b+a+c}{2}=\frac{2a+b+c}{2}\)

hay \(\sqrt{a^2+1}\leq \frac{2a+b+c}{2}\)

Hoàn toàn tương tự với các biểu thức còn lại và cộng theo vế:

\(\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}\leq \frac{2a+b+c}{2}+\frac{2b+a+c}{2}+\frac{2c+a+b}{2}=2(a+b+c)\)

Ta có đpcm. Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)

Bài 4:

Ta có:

\(A=\frac{8a^2+b}{4a}+b^2=2a+\frac{b}{4a}+b^2\)

\(\Leftrightarrow A+\frac{1}{4}=2a+\frac{b+a}{4a}+b^2=2a+b+\frac{b+a}{4a}+b^2-b\)

Vì \(a+b\geq 1, a>0\) nên \(A+\frac{1}{4}\geq a+1+\frac{1}{4a}+b^2-b\)

Áp dụng BĐT AM-GM:

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

\(\Rightarrow A+\frac{1}{4}\geq 2+b^2-b=\left(b-\frac{1}{2}\right)^2+\frac{7}{4}\geq \frac{7}{4}\)

\(\Leftrightarrow A\geq \frac{3}{2}\).

Vậy \(A_{\min}=\frac{3}{2}\Leftrightarrow a=b=\frac{1}{2}\)

Bài 5:

Ta có : \(\sqrt{x+2}-y^3=\sqrt{y+2}-x^3\)

\(\Leftrightarrow (\sqrt{x+2}-\sqrt{y+2})+(x^3-y^3)=0\)

\(\Leftrightarrow (x-y)\left(\frac{1}{\sqrt{x+2}+\sqrt{y+2}}+x^2+xy+y^2\right)=0\)

Ta thấy biểu thức trong ngoặc lớn luôn lớn hơn $0$ , do đó \(x-y=0\Leftrightarrow x=y\)

Khi đó, \(A=x^2+2x+10=(x+1)^2+9\geq 9\)

Vậy \(A_{\min}=9\Leftrightarrow x=y=-1\)

Ta có \(2=a^2+b^2\ge2ab\)

\(\Leftrightarrow ab\le1\)

\(M\le\sqrt{\left(a^2+b^2\right)\left(36ab+45b^2+36ab+45a^2\right)}\)

\(=\sqrt{2\left(72ab+90\right)}\)\(\le\sqrt{2\left(72+90\right)}=\sqrt{324}=18\)

GTLN là 18 đạt được khi a = b = 1

\(\sqrt{a\left(8a+b\right)}=\dfrac{1}{3}\cdot\sqrt{9a\left(8a+b\right)}< =\dfrac{1}{3}\cdot\dfrac{9a+8a+b}{2}=\dfrac{1}{6}\left(17a+b\right)\)

\(\sqrt{b\left(8b+a\right)}< =\dfrac{1}{6}\left(17b+a\right)\)

=>\(\sqrt{a\left(8a+b\right)}+\sqrt{b\left(8b+a\right)}< =3\left(a+b\right)\)

=>\(\dfrac{a+b}{\sqrt{a\left(8a+b\right)}+\sqrt{b\left(8b+a\right)}}>=\dfrac{1}{3}\)

Áp dụng BĐT Cô-si cho 2 số không âm

Ta có: \(\sqrt{9b\left(4a+b\right)}\)\(\le\) \(\dfrac{9b+4a+5b}{2}\)=\(\dfrac{14b+4a}{2}\)

\(\Rightarrow\) \(a\sqrt{9b\left(4a+5b\right)}\)\(\le\) \(\dfrac{14ab+4a^2}{2}\)=7ab+2a²

^{CMTT: \(b\sqrt{9a\left(4b+5a\right)}\)} \(\le\) 7ab+2b²

^{\(\Rightarrow\)} M\(\le\) 14ab + 2(a²+b²) \(\le\)7(a²+b²) + 2(a²+b²) = 9(a²+b²)=18

Vậy M_min=18

Dấu "=" xảy ra\(\Leftrightarrow\) a=b=1

\(M=a\sqrt{9b\left(4a+5b\right)}+b\sqrt{9a\left(4b+5a\right)}\le\dfrac{a\left(9b+4a+5b\right)}{2}+\dfrac{b\left(9a+4b+5a\right)}{2}=\dfrac{a\left(14b+4a\right)+b\left(14a+4b\right)}{2}=2a^2+7ab+7ab+2b^2=2\left(a^2+b^2\right)+14ab=4+14ab\le4+14\times\dfrac{a^2+b^2}{2}=4+14=18\)

Dấu "=" xảy ra <=> a = b = 1