Đáp án:
$GTNN_P=\dfrac{17}{4}↔a=b=\dfrac{1}{2}$
Giải thích các bước giải:
$\begin{array}{l}P=\dfrac{a^4+1}{a}+\dfrac{b^4+1}{b}\\P=\dfrac{a^4}{a}+\dfrac{1}{a}+\dfrac{b^4}{b}+\dfrac{1}{b}\\P=a^3+b^3+\dfrac{1}{a}+\dfrac{1}{b}\\\text{áp dụng BĐT svacxơ ta có}\\\dfrac{1}{a}+\dfrac{1}{b} \geq \dfrac{4}{a+b}=4(a+b=1)\\→P \geq a^3+b^3+4\\↔P \geq 4+(a+b)(a^2-ab+b^2)\\↔P \geq 4+a^2-ab+b^2\\↔P \geq 4+a^2+2ab+b^2-3ab\\↔P \geq 4+(a+b)^2-3ab\\↔P \geq 4+1-3ab\\↔P \geq 5-3ab\\\text{áp dụng BĐT cosi ta có}\\a^2+b^2 \geq 2ab\\↔a^2+2ab+b^2 \geq 4ab\\↔4ab \leq (a+b)^2=1\\↔ab \leq \dfrac{1}{4}\\↔P \geq 5-\dfrac{3}{4}\\↔P \geq \dfrac{17}{4}\\\text{dấu = xảy ra khi}\\\begin{cases}a=b\\a+b=1\\\end{cases}
\\↔a=b=\dfrac{1}{2}\\vậy \,\, GTNN_P=\dfrac{17}{4}↔a=b=\dfrac{1}{2}\\\underline{\text{CHÚC BẠN HỌC TỐT}}\\\end{array}$
