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 AM - GM:

\(P=3a+3b-1+\left[\left(a+1\right)+b+\dfrac{c^3}{b\left(a+1\right)}\right]\ge3a+3b-1+3c=3.5-1=14\).

Đẳng thức xảy ra khi a = 1; b = 2; c = 2.

Vậy Min P = 14 khi a = 1; b = 2; c = 2.

Lời giải:

$a^2-2ab-3b^2\geq 0$

$\Leftrightarrow (a^2+ab)-(3ab+3b^2)\geq 0$

$\Leftrightarrow a(a+b)-3b(a+b)\geq 0$

$\Leftrightarrow (a+b)(a-3b)\geq 0$

$\Leftrightarrow a-3b\geq 0$ (do $a+b>0$ với mọi $a,b>0$)

$\Leftrightarrow a\geq 3b$

Xét hiệu:

$P-\frac{37}{3}=\frac{4a^2+b^2}{ab}-\frac{37}{3}$

$=\frac{12a^2+3b^2-37ab}{3ab}=\frac{(a-3b)(12a-b)}{3ab}\geq 0$ do $a\geq 3b>0$

$\Rightarrow P\geq \frac{37}{3}$

Vậy $P_{\min}=\frac{37}{3}$

Lời giải:
Áp dụng BĐT Cô-si:

$a^2+1\geq 2a$

$b^2+4\geq 4b$

$\Rightarrow a^2+b^2\geq 2a+4b-5$

$\Rightarrow P\geq 2a+4b-5+\frac{1}{a+b}+\frac{1}{b}$

$=\frac{a+b}{9}+\frac{1}{a+b}+(\frac{b}{4}+\frac{1}{b})+\frac{17}{9}a+\frac{131}{36}b-5$

$\geq 2\sqrt{\frac{1}{9}}+2\sqrt{\frac{1}{4}}+\frac{17}{9}a+\frac{131}{36}b-5$

$=\frac{2}{3}+1+\frac{17}{9}a+\frac{131}{36}b-5$

$\geq \frac{2}{3}+1+\frac{17}{9}+\frac{131}{36}.2-5=\frac{35}{6}$

Vậy $P_{\min}=\frac{35}{6}$ khi $a=1; b=2$

Ta thấy \(ab\le\dfrac{a^2+b^2}{2}=1\) và \(a+b\le\sqrt{2\left(a^2+b^2\right)}=2\). Áp dụng BĐT B.C.S, ta được \(P=\dfrac{a^4}{ba^2+a^2}+\dfrac{b^4}{ab^2+b^2}\) \(\ge\dfrac{\left(a^2+b^2\right)^2}{ba^2+ab^2+a^2+b^2}=\dfrac{2^2}{ab\left(a+b\right)+2}\ge\dfrac{4}{1.2+2}=1\)

ĐTXR \(\Leftrightarrow a=b=1\)

Vậy GTNN của P là 1 khi \(a=b=1\)

\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)

\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)

\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

https://hoc24.vn/cau-hoi/cho-abc-0-thoa-man-abbcca3-tim-gia-tri-nho-nhat-cua-pdfrac13a1b2dfrac13b1c2dfrac13c1a2.6181078378966

Từ giả thiết \(1\le a\le2\) =>  ( a - 1).(a - 2) \(\le\) 0 =>\(a^2-3a+2\le0\)

Từ giả thiết \(1\le b\le2\) => (b - 1)( b - 2) \(\le\) 0 => \(a^2-3b+2\le0\)

Vì vậy ta có P:

\(=\left[a^2+b^2-3\left(a+b\right)+4\right]-\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)^2-\left(\dfrac{\sqrt{b}}{2}-\dfrac{1}{\sqrt{b}}\right)^2-3\le-3\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{a}=\dfrac{1}{\sqrt{q}}\\\dfrac{\sqrt{b}}{2}=\dfrac{1}{\sqrt{b}}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)

Vậy a =1 ; b = 2 là giá trị lớn nhất của biểu thức