1. ĐKXĐ: ...
Đặt \(2\sqrt{x+2}+\sqrt{4x+1}=t\ge\sqrt{7}\)
\(\Rightarrow t^2=8x+9+4\sqrt{4x^2+9x+2}\)
\(\Rightarrow2x+\sqrt{4x^2+9x+2}=\frac{t^2-9}{4}\)
Phương trình trở thành:
\(\frac{t^2-9}{4}+3=t\)
\(\Leftrightarrow t^2-4t+3=0\Rightarrow\left[{}\begin{matrix}t=1\left(l\right)\\t=3\end{matrix}\right.\)
\(\Rightarrow4\sqrt{4x^2+9x+2}=t^2-\left(8x+9\right)=-8x\) (\(x\le0\))
\(\Leftrightarrow\sqrt{4x^2+9x+2}=-2x\)
\(\Leftrightarrow4x^2+9x+2=4x^2\Rightarrow x=-\frac{2}{9}\)
Bài 2:
Ta có: \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}\Rightarrow3\ge a+b+c\)
Do \(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\sqrt{a}+\sqrt{b}+\sqrt{c}\)
Nên BĐT sẽ được chứng minh nếu ta chỉ ra rằng:
\(\sqrt{a}+\sqrt{b}+\sqrt{c}\ge ab+bc+ca\)
Thật vậy, ta có:
\(\sqrt{a}+\sqrt{a}+a^2\ge3a\) ; \(\sqrt{b}+\sqrt{b}+b^2\ge3b\) ; \(\sqrt{c}+\sqrt{c}+c^2\ge3c\)
\(\Rightarrow2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+a^2+b^2+c^2\ge3\left(a+b+c\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+a^2+b^2+c^2\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}\ge ab+bc+ca\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
Bài 2 :
Ta có\(VT=\frac{a^2}{a\sqrt{b}}+\frac{b^2}{b\sqrt{c}}+\frac{c^2}{c\sqrt{a}}\ge\frac{\left(a+b+c\right)^2}{a\sqrt{b}+b\sqrt{c}+c\sqrt{a}}\)
Mà \(a\sqrt{b}\le\frac{a^2+b}{2},b\sqrt{c}\le\frac{b^2+c}{2},c\sqrt{a}\le\frac{c^2+a}{2}\)
\(\Rightarrow VT\ge\frac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+a+b+c}=\frac{2\left(a+b+c\right)^2}{3+a+b+c}\)
Lại có : \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)
\(\Rightarrow VT\ge\frac{2\left(a+b+c\right)^2}{3+3}=\frac{\left(a+b+c\right)^2}{3}\ge ab+bc+ac\left(đpcm\right)\)
Dấu " = " xảy ra khi a=b=c=1
