Xét \(a+b\ge1\Leftrightarrow b\ge1-a\)

Xét \(Q\ge\dfrac{8a^2+1-a}{4a}+\left(1-a\right)^2=\dfrac{8a^2}{4a}+\dfrac{1}{4a}-\dfrac{a}{4a}+1-2a+a^2\)

        \(=2a+\dfrac{1}{4a}-\dfrac{1}{4}+1-2a+a^2\)\(=a^2+\dfrac{1}{4a}+\dfrac{3}{4}\)\(=\left(a^2+\dfrac{1}{8a}+\dfrac{1}{8a}\right)+\dfrac{3}{4}\)

Áp dụng Cosi được \(Q\ge3\sqrt[3]{a^2\cdot\dfrac{1}{8a}\cdot\dfrac{1}{8a}}+\dfrac{3}{4}\)\(=3\sqrt[3]{\dfrac{1}{64}}+\dfrac{3}{4}=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\) 

Vậy \(Qmin=\dfrac{3}{2}\) khi \(a=b=\dfrac{1}{2}\)

Từ \(a+b\ge1=>b\ge1-a>0\) ta có:

A = \(\dfrac{8a^2+b}{4a}+b^2\ge\dfrac{8a^2+1-a}{4a}+\left(1-a\right)^2\)

=\(\dfrac{8a^2-a+1+4a^3-8a^2+4a}{4a}=\dfrac{4a^3-4a^2+a+4a^2-4a+1+6a}{4a}\)

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

Vì với a>0 thì\(\dfrac{\left(2a-1\right)^2\left(a+1\right)}{4a}\ge0\)

Dấu = xảy ra khi a=1/2

Nên từ (1) => A\(\ge0+\dfrac{3}{2}\) hay A\(\ge\dfrac{3}{2}\)

Vậy GTNN của A=3/2 khi a=b=1/2

A = \(\dfrac{8a^2+b}{4a}+b^2\)

Ta có: a + b \(\ge\) 1 \(\Leftrightarrow\) b \(\ge\) 1 - a

\(\Rightarrow\) A \(\ge\) \(\dfrac{8a^2+1-a}{4a}+\left(1-a\right)^2\)

\(\Leftrightarrow\) A \(\ge\) 2a + \(\dfrac{1}{4a}\) - \(\dfrac{1}{4}\) + 1 - 2a + a²

\(\Leftrightarrow\) A \(\ge\) a² + \(\dfrac{1}{4a}\) + \(\dfrac{3}{4}\)

\(\Leftrightarrow\) A \(\ge\) a² + \(\dfrac{1}{8a}\) + \(\dfrac{1}{8a}\) + \(\dfrac{3}{4}\)

Áp dụng BĐT Cô-si cho 3 số dương a²; \(\dfrac{1}{8a}\); \(\dfrac{1}{8a}\)

a² + \(\dfrac{1}{8a}\) + \(\dfrac{1}{8a}\) \(\ge\) 3\(\sqrt[3]{\dfrac{a^2}{64a^2}}\) = 3\(\sqrt[3]{64}\) = 3.4 = 12

\(\Leftrightarrow\) a² + \(\dfrac{1}{8a}\) + \(\dfrac{1}{8a}\) + \(\dfrac{3}{4}\) \(\ge\) 12 + \(\dfrac{3}{4}\) = \(\dfrac{51}{4}\)

Hay A \(\ge\) a² + \(\dfrac{1}{8a}\) + \(\dfrac{1}{8a}\) + \(\dfrac{3}{4}\) \(\ge\) \(\dfrac{51}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\) a² = \(\dfrac{1}{8a}\) \(\Leftrightarrow\) 8a³ = 1 \(\Leftrightarrow\) a³ = \(\dfrac{1}{8}\) \(\Leftrightarrow\) a = \(\dfrac{1}{2}\)

và b = 1 - a \(\Leftrightarrow\) b = 1 - \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\)

Vậy Min_A = \(\dfrac{51}{4}\) \(\Leftrightarrow\) a = b = \(\dfrac{1}{2}\)

 Chúc bn học tốt! (ko chắc lắm đâu)

Áp dụng BĐT Bunyakovsky, ta có:

\(a+b+c\le\sqrt{3(a^2+b^2+c^2)}=\sqrt{3.3}=3\)

Áp dụng BĐT Cauchy, ta có:

\(A=\sum{\dfrac{1}{\sqrt{1+8a^3}}}=\sum{\dfrac{1}{\sqrt{(2a+1)(4a^2-2a+1)}}} \\\ge\sum{\dfrac{1}{\dfrac{4a^2+2}{2}}}=\sum{\dfrac{1}{2a^2+1}} \)

Ta cần chứng minh: \(\dfrac{1}{2a^2+1}\ge\dfrac{-4}{9}a+\dfrac{7}{9} \\<=>\dfrac{8a^3-14a^2+4a+2}{9(2a^2+1)}\ge0 \\<=>\dfrac{2(a-1)^2(4a+1)}{9(2a^2+1)}\ge0 (luôn\ đúng\ với\ mọi\ a>0) \\->\sum{\dfrac{1}{2a^2+1}}\ge\dfrac{-4}{9}(a+b+c)+\dfrac{21}{9}\ge\dfrac{-4}{9}.3+\dfrac{21}{9}=1 \\->A\ge1 \)

Đẳng thức xảy ra khi a = b = c = 1.

Vậy GTNN của A là 1 (khi a = b = c = 1).

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

\(\ge a+1-b+\frac{1-a}{4a}+b^2=a+1-b+\frac{1}{4a}-\frac{1}{4}+b^2\)(do \(a+b\ge1\))

\(=\left(a+\frac{1}{4a}\right)+b^2-b+\frac{1}{4}+\frac{1}{2}\)

\(\ge2\sqrt{a\cdot\frac{1}{4a}}+\left(b-\frac{1}{2}\right)^2+\frac{1}{2}\)

\(\ge2\cdot\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\)

Dấu = khi \(a=b=\frac{1}{2}\)

Lời giải:

Áp dụng BĐT AM-GM:

$1\geq a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{1}{4}$

Áp dụng BĐT Cauchy-Schwarz:
\(A=\frac{1}{1+a^2+b^2}+\frac{1}{6ab}+\frac{1}{3ab}\geq \frac{4}{1+a^2+b^2+6ab}+\frac{1}{3ab}\)

\(=\frac{4}{1+(a+b)^2+4ab}+\frac{1}{3ab}\geq \frac{4}{1+1+4.\frac{1}{4}}+\frac{1}{3.\frac{1}{4}}=\frac{8}{3}\)

Vậy $A_{\min}=\frac{8}{3}$ khi $a=b=\frac{1}{2}$