Đáp án:

Giải thích các bước giải:

Thử áp dụng Cô si cho 3 số ( nếu đã học qua):

$ \dfrac{b}{4a} + \dfrac{c}{4b} + \dfrac{a}{4c} ≥ 3\sqrt[3]{\dfrac{b}{4a}.\dfrac{c}{4b}.\dfrac{a}{4c}} = \dfrac{3}{4} $ 

$ ⇔ (a + b + c) - (\dfrac{b}{4a} + \dfrac{c}{4b} + \dfrac{a}{4c}) ≤  \dfrac{3}{2} - \dfrac{3}{4}$ 

$ ⇔ \dfrac{4a² - b}{4a} + \dfrac{4b² - c}{4b} + \dfrac{4c² - a}{4c} ≤ \dfrac{3}{4} (*)$

Do $ (2x - 1)² ≥ 0 ⇔ 4x² + 1 ≥ 4x$ với $∀x$ và $ a, b, c > 0$ nên từ $(*)$

$ ⇒ \dfrac{4a² - b}{1 + 4a²} + \dfrac{4b² - c}{1 + 4b²} + \dfrac{4c² - a}{1 + 4c²} ≤ \dfrac{3}{4}$

$ ⇔ (1 - \dfrac{4a² - b}{1 + 4a²}) + (1 - \dfrac{4b² - c}{1 + 4b²}) + (1 - \dfrac{4c² - a}{1 + 4c²}) ≥ 3 - \dfrac{3}{4}$

$ ⇔ \dfrac{1 + b}{1 + 4a²} + \dfrac{1 + c}{1 + 4b²} + \dfrac{1 + a}{1 + 4c²}  ≥ \dfrac{9}{4}$

$ ⇒ MinP = \dfrac{9}{4} ⇔ a = b = c = \dfrac{1}{2}$

Đáp án:

The minimum value of $P$ is $\dfrac94\Leftrightarrow a = b = c =\dfrac32$

Giải thích các bước giải:

$P = \dfrac{1+b}{1+ 4a^2} +\dfrac{1+c}{1 + 4b^2} +\dfrac{1+a}{1 + 4c^2}$

$\Leftrightarrow P =\dfrac{b}{1 + 4a^2} +\dfrac{c}{1 + 4b^2} +\dfrac{a}{1+ 4c^2} +\dfrac{1}{1 + 4a^2} +\dfrac{1}{1 + 4b^2} +\dfrac{1}{1+ 4c^2}$

$\Leftrightarrow P = \left(b -\dfrac{4a^2b}{1 + 4a^2}\right) + \left(c -\dfrac{4b^2c}{1 + 4b^2}\right) + \left(a -\dfrac{4c^2a}{1 + 4c^2}\right) + \left(1 -\dfrac{4a^2}{1 + 4a^2}\right) + \left(1 -\dfrac{4b^2}{1 + 4b^2}\right) + \left(1 -\dfrac{4c^2}{1 + 4c^2}\right)$

$\Leftrightarrow P = (a + b + c) -\left(\dfrac{4a^2b}{1 + 4a^2} + \dfrac{4b^2c}{1 + 4b^2}+\dfrac{4c^2a}{1 + 4c^2}\right) + 3 -\left(\dfrac{4a^2}{1 + 4a^2} +\dfrac{4b^2}{1 + 4b^2}+ \dfrac{4c^2}{1 + 4c^2}\right)$

By using $AM-GM$ inequality, we have:

$1 + 4a^2 \geq 2\sqrt{4a^2}= 4a$

$1 + 4b^2 \geq 4b$

$1 + 4c^2 \geq 4c$

Therefore:

$\left(\dfrac{4a^2b}{1 + 4a^2} + \dfrac{4b^2c}{1 + 4b^2}+\dfrac{4c^2a}{1 + 4c^2}\right)+ \left(\dfrac{4a^2}{1 + 4a^2} +\dfrac{4b^2}{1 + 4b^2}+ \dfrac{4c^2}{1 + 4c^2}\right)$

$\leq \left(\dfrac{4a^2b}{4a} +\dfrac{4b^2c}{4b} +\dfrac{4c^2a}{4c}\right)+\left(\dfrac{4a^2}{4a} +\dfrac{4b^2}{4b} +\dfrac{4c^2}{4c}\right)$

$\Leftrightarrow- \left(\dfrac{4a^2b}{1 + 4a^2} + \dfrac{4b^2c}{1 + 4b^2}+\dfrac{4c^2a}{1 + 4c^2}\right)-\left(\dfrac{4a^2}{1 + 4a^2} +\dfrac{4b^2}{1 + 4b^2}+ \dfrac{4c^2}{1 + 4c^2}\right)$

$\geq -(ab +bc + ca) - (a + b + c)$

$\Leftrightarrow (a + b + c) -\left(\dfrac{4a^2b}{1 + 4a^2} + \dfrac{4b^2c}{1 + 4b^2}+\dfrac{4c^2a}{1 + 4c^2}\right) + 3 -\left(\dfrac{4a^2}{1 + 4a^2} +\dfrac{4b^2}{1 + 4b^2}+ \dfrac{4c^2}{1 + 4c^2}\right)$

$\geq (a+b+c)-(ab +bc + ca) +3- (a+b+c)$

$\Leftrightarrow P \geq 3 - (ab + bc + ca)$

Beside, we also have:

$(a + b + c)^2 \geq 3(ab + bc+ ca)$

$\Leftrightarrow (ab + bc + ca)\leq \dfrac{(a+b+c)^2}{3}$

$\Leftrightarrow -(ab+bc + ca)\geq -\dfrac{(a+b+c)^2}{3}$

$\Leftrightarrow 3 -(ab+bc + ca)\geq 3-\dfrac{(a+b+c)^2}{3}$

$\Leftrightarrow P\geq 3 -\left(\dfrac32\right)^2\cdot\dfrac13= \dfrac94$

Equality holds if and only if $a = b = c =\dfrac12$

So the minimum value of $P$ is $\dfrac94\Leftrightarrow a = b = c =\dfrac32$