Question

[Khiêm Nguyễn Gia]

Cho bốn số thực dương \(a,b,x,y\) thỏa mãn \(a+b=4ab\). Chứng minh rằng:
\(a\)) \(\dfrac{a^2}{x}+\dfrac{b^2}{y}\ge\dfrac{\left(a+b\right)^2}{x+y}\).
\(b\)) \(\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}\ge\dfrac{1}{2}\).

Answer 1

[Lê Song Phương]

a) BĐT cần chứng minh \(\Leftrightarrow\dfrac{a^2y+b^2x}{xy}\ge\dfrac{a^2+2ab+b^2}{x+y}\)

\(\Leftrightarrow a^2xy+a^2y^2+b^2x^2+b^2xy\ge xya^2+2abxy+xyb^2\)

\(\Leftrightarrow a^2y^2-2abxy+b^2x^2\ge0\)

\(\Leftrightarrow\left(ay-bx\right)^2\ge0\) (luôn đúng)

Vậy ta có đpcm. Dấu "=" xảy ra khi \(ay=bx\)

b) Ta có \(VT=\dfrac{a^2}{4b^2a+a}+\dfrac{b^2}{4a^2b+b}\)

\(\ge\dfrac{\left(a+b\right)^2}{4ab\left(a+b\right)+\left(a+b\right)}\)

\(=\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2+a+b}\) (vì \(4ab=a+b\))

\(=\dfrac{a+b}{a+b+1}\)

Đặt \(t=a+b\left(t>0\right)\) thì suy ra \(VT\ge\dfrac{t}{t+1}\)

Do \(4ab=a+b\ge2\sqrt{ab}\Leftrightarrow ab\ge\dfrac{1}{4}\)

Nên \(a+b\ge1\) \(\Rightarrow t\ge1\)

Ta cần tìm GTNN của \(T=\dfrac{t}{t+1}\) với \(t\ge1\)

\(T=\dfrac{1}{1+\dfrac{1}{t}}\)

Ta có \(t\ge1\Leftrightarrow\dfrac{1}{t}\le1\Leftrightarrow1+\dfrac{1}{t}\le2\Leftrightarrow\dfrac{1}{1+\dfrac{1}{t}}\ge\dfrac{1}{2}\)

Vậy \(T\ge\dfrac{1}{2}\) \(\Leftrightarrow VT\ge\dfrac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\dfrac{a}{4b^2a+a}=\dfrac{b}{4a^2b+b}\) và \(t=1\)

\(\Leftrightarrow4a^3b+ab=4b^3a+ab\) và \(a+b=1\)

\(\Leftrightarrow a=b\) và \(a+b=1\)

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

Vậy ta có đpcm. Dấu "=" xảy ra \(\Leftrightarrow a=b=\dfrac{1}{2}\)