Bt=4/2ab+3/(a^2+b^2)=1/2ab+3(1/2ab+1/a^2+b^2)

>=1/2ab+3.4/(a+b)^2(BĐT Cauchuy-Swartch)

>=2/4ab+12/(a+b)^2>=2(a+b)^2+12/(a+b)^2=14/(a+b)^2=1

Dấu= xảy ra khi a=b=1/2

ab là 1/2

a/b là 1/2

\(ab+bc+ac=3\)

Ta có:

\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\) ( đúng với mọi \(ab\ge1\))

Giả sử:\(ab\ge1\)

\(\Rightarrow\dfrac{2}{ab+1}+\dfrac{1}{c^2+1}\ge\dfrac{2c^2+2+ab+1}{\left(ab+1\right)\left(c^2+1\right)}=\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\)

Giả sử: \(\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\ge\dfrac{3}{2}\)(đúng)

\(\Leftrightarrow2\left(2c^2+ab+3\right)\ge3\left(ab+1\right)\left(c^2+1\right)\)

\(\Leftrightarrow4c^2+2ab+6\ge3\left(abc^2+ab+c^2+1\right)\)

\(\Leftrightarrow4c^2+2ab+6\ge3abc^2+3ab+3c^2+3\)

\(\Leftrightarrow c^2-ab-3abc^2+3\ge0\)

\(\Leftrightarrow c^2-ab-3abc^2+ab+ac+bc\ge0\)   ( vì \(ab+ac+bc=3\) )

\(\Leftrightarrow c^2+ac+bc-3abc^2\ge0\)

\(\Leftrightarrow c+a+b-3abc\ge0\)

\(\Leftrightarrow c+a+b\ge3abc\)

Ta có:

\(3\left(c+a+b\right)=\left(ab+ac+bc\right)\left(c+a+b\right)\) ( vì \(ab+ac+bc=3\) )

Áp dụng BĐT AM-GM, ta có:

\(\left(ab+ac+bc\right)\left(c+a+b\right)\ge9abc\)

\(\Rightarrow a+b+c\ge3abc\)

\(\Rightarrow\) \(\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\ge\dfrac{3}{2}\) ( luôn đúng )

\(\Rightarrow\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{3}{2}\) ( đfcm )

Dấu "=" xảy ra khi \(a=b=c=1\)

Hình như sai đề rồi bạn ạ, dấu ≥ phải là ≤

Tham khảo: https://lazi.vn/edu/exercise/cho-a-b-c-la-cac-so-duong-thoa-man-a2-2b2-3c2-chung-minh-1-a-2-b-3-c

\(VT=3\left(\dfrac{1}{4ab}+\dfrac{1}{a^2+4b^2}\right)+\dfrac{1}{2.a.2b}\ge\dfrac{12}{a^2+4ab+4b^2}+\dfrac{2}{\left(a+2b\right)^2}=14\)

Dấu "=" xảy ra khi \(\left(a;b\right)=\left(\dfrac{1}{2};\dfrac{1}{4}\right)\)

Ta có \(-\dfrac{4ab^2}{4b^2+1}\ge-\dfrac{4ab^2}{2\sqrt{4b^2}}=\dfrac{4ab^2}{4b}=ab\)

\(-\dfrac{4a^2b}{4a^2+1}\ge-\dfrac{4a^2b}{2\sqrt{4a^2}}=\dfrac{4a^2b}{4a}=ab\)

Mà \(\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}=\dfrac{a\left(4b^2+1\right)}{4b^2+1}-\dfrac{4ab^2}{4b^2+1}+\dfrac{b\left(4a^2+1\right)}{4a^2+1}-\dfrac{4ab^2}{4a^2+1}\ge a-ab+b-ab=4ab-2ab=2ab\)

Mà \(a+b=4ab\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=4\ge\dfrac{2}{2\sqrt{ab}}\Rightarrow4\sqrt{ab}\ge2\Rightarrow ab\ge\dfrac{1}{4}\)

\(\Rightarrow2ab\ge\dfrac{1}{2}\Rightarrow\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}\ge\dfrac{1}{2}\)

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

Lời giải:

ĐK $\Rightarrow \frac{1}{a}+\frac{1}{b}=4$

Đặt $\frac{1}{x}=a; \frac{1}{y}=b$ thì bài toán trở thành:

Cho $a,b>0$ thỏa mãn $a+b=4$. CMR:

$P=\frac{x^2}{y(x^2+4)}+\frac{y^2}{x(y^2+4)}\geq \frac{1}{2}$

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Áp dụng BĐT AM-GM:

$\frac{x^2}{y(x^2+4)}+\frac{y(x^2+4)}{64}\geq \frac{x}{4}$

$\frac{y^2}{x(y^2+4)}+\frac{x(y^2+4)}{64}\geq \frac{y}{4}$

Cộng theo vế và rút gọn:

$P\geq \frac{3(x+y)-xy}{16}=\frac{12-xy}{16}$

Mà $xy\leq \frac{(x+y)^2}{4}=4$

$\Rightarrow P\geq \frac{12-4}{16}=\frac{1}{2}$

Ta có đpcm.