Đề sai với a = b = 1.

BĐT cần chứng minh tương đương với:

\(\left(\dfrac{a}{b^2}-\dfrac{2}{b}+\dfrac{1}{a}\right)+\left(\dfrac{b}{a^2}-\dfrac{2}{a}+\dfrac{1}{b}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{16}{a+b}\)

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

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

\(\Leftrightarrow\left(a-b\right)^2\left[\dfrac{a+b}{a^2b^2}-\dfrac{4}{ab\left(a+b\right)}\right]\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)^4}{a^2b^2\left(a+b\right)}\ge0\) (luôn đúng).

`a/b^2+b/a^2+16/(a+b)>=5(1/a+1/b)`

`<=>a/b^2-1/b+b^2-1/a+4(4/(a+b)-1/a-1/b)=0`

`<=>(a-b)/b^2+(b-a)/a^2+4((4ab-(a+b)^2)/(ab(a+b)))>=0`

`<=>(a^2(a-b)-b^2(a-b))/(a^2b^2)-(4(a-b)^2)/(ab(a+b))>=0`

`<=>(a-b)^2[(a+b)^2-4ab]>=0`

`<=>(a-b)^2(a^2-2ab+b^2)>=0`

`<=>(a-b)^2(a-b)^2>=0`

`<=>(a-b)^4>=0` luôn đúng.

Dấu "=" xảy ra khi `a=b`

Áp dụng BĐT AM-GM ta có:

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

\(=\left(a+\dfrac{1}{4a}\right)+\left(b+\dfrac{1}{4b}\right)+3\left(\dfrac{1}{4a}+\dfrac{1}{4b}\right)\)

\(=2\sqrt{a\cdot\dfrac{1}{4a}}+2\sqrt{b\cdot\dfrac{1}{4b}}+3\dfrac{\left(1+1\right)^2}{4\left(a+b\right)}\)

\(\ge2\cdot\dfrac{1}{2}+2\cdot\dfrac{1}{2}+\dfrac{3\cdot4}{4}=5=VP\)

Xảy ra khi \(a=b=\dfrac{1}{2}\)

Lời giải:

Dùng pp biến đổi tương đương.

Ta có: \(\frac{a}{b^2}+\frac{b}{a^2}+\frac{16}{a+b}\geq 5\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\Leftrightarrow \left(\frac{a}{b^2}-\frac{1}{b}\right)+\left(\frac{b}{a^2}-\frac{1}{a}\right)+4\left(\frac{4}{a+b}-\frac{1}{a}-\frac{1}{b}\right)\geq 0\)

\(\Leftrightarrow \frac{a-b}{b^2}-\frac{a-b}{a^2}+4\left(\frac{4}{a+b}-\frac{a+b}{ab}\right)\geq 0\)

\(\Leftrightarrow \frac{(a-b)^2(a+b)}{a^2b^2}-\frac{4(a-b)^2}{ab(a+b)}\geq 0\)

\(\Leftrightarrow (a-b)^2\left(\frac{a+b}{a^2b^2}-\frac{4}{ab(a+b)}\right)\geq 0\)

\(\Leftrightarrow \frac{a+b}{a^2b^2}-\frac{4}{ab(a+b)}\geq 0\)

\(\Leftrightarrow \frac{a+b}{ab}-\frac{4}{a+b}\geq 0\Leftrightarrow (a+b)^2-4ab\geq 0\)

\(\Leftrightarrow (a-b)^2\geq 0\) (luôn đúng)

Do đó ta có đpcm.

Dấu bằng xảy ra khi $a=b$

Bài này bị ngược dấu hả ???

Đây nhé , ta sẽ chứng minh \(\frac{a}{b^2}+\frac{b}{a^2}\ge\frac{1}{a}+\frac{1}{b}\) thật vậy

Áp dụng bđt Cô-si cho 2 số dương ta được

\(\frac{a}{b^2}+\frac{1}{a}\ge2\sqrt{\frac{a}{b^2}.\frac{1}{a}}=\frac{2}{b}\)

\(\frac{b}{a^2}+\frac{1}{b}\ge\frac{2}{b}\)

Cộng 2 bđt lại ta được \(\frac{a}{b^2}+\frac{b}{a^2}+\frac{1}{a}+\frac{1}{b}\ge2\left(\frac{1}{a}+\frac{1}{b}\right)\)

                                 \(\Leftrightarrow\frac{a}{b^2}+\frac{b}{a^2}\ge\frac{1}{a}+\frac{1}{b}\)

Dấu ''=" xảy ra khi a = b

Bài toán quay trở lại với việc c/m \(\frac{16}{a+b}\ge\frac{4}{a}+\frac{4}{b}\)với a,b > 0 

Ta có bđt sau \(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)(Quen thuộc)

Áp dụng ta được \(\frac{4}{a}+\frac{4}{b}=\frac{2^2}{a}+\frac{2^2}{b}\ge\frac{\left(2+2\right)^2}{a+b}=\frac{16}{a+b}\)

\(\Rightarrow\frac{4}{a}+\frac{4}{b}\ge\frac{16}{a+b}???\) Trái với điều cần c/m

=> Đề sai 

\(\frac{a}{b^2}+\frac{b}{a^2}+\frac{16}{a+b}\ge5.\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\Leftrightarrow\frac{a^3+b^3}{a^2b^2}+\frac{16}{a+b}\ge\frac{5.\left(a+b\right)}{ab}\)

\(\Leftrightarrow\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2b^2}+\frac{16}{a+b}\ge\frac{5.\left(a+b\right)}{ab}\)

\(\Leftrightarrow\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{ab}+\frac{16ab}{a+b}\ge5.\left(a+b\right)\)

\(\Leftrightarrow\frac{a^2-ab+b^2}{ab}+\frac{16ab}{\left(a+b\right)^2}\ge5\)

\(\Leftrightarrow\frac{a^2+b^2}{ab}+\frac{16ab}{\left(a+b\right)^2}-1\ge5\)

\(\Leftrightarrow\frac{a^2+b^2}{ab}+\frac{16ab}{\left(a+b\right)^2}\ge6\)

\(\Leftrightarrow\frac{\left(a+b\right)^2}{ab}+\frac{16ab}{\left(a+b\right)^2}-2\ge6\)

\(\Leftrightarrow\frac{\left(a+b\right)^2}{ab}+\frac{16ab}{\left(a+b\right)^2}\ge8\) (1)

Áp dụng BĐT AM-GM ta có:

\(\frac{\left(a+b\right)^2}{ab}+\frac{16ab}{\left(a+b\right)^2}\ge2.\sqrt{\frac{\left(a+b\right)^2}{ab}.\frac{16ab}{\left(a+b\right)^2}}=2.\sqrt{16}=2.4=8\)(2)

Từ (1) và (2)

\(\Rightarrow\frac{a}{b^2}+\frac{b}{a^2}+\frac{16}{a+b}\ge5.\left(\frac{1}{a}+\frac{1}{b}\right)\)

Dấu " = " xảy ra \(\Leftrightarrow\frac{\left(a+b\right)^2}{ab}=\frac{16ab}{\left(a+b\right)^2}\Leftrightarrow\left(a+b\right)^4=\left(4ab\right)^2\Leftrightarrow a^2+2ab+b^2=4ab\Leftrightarrow a^2-2ab+b^2=0\Leftrightarrow\left(a-b\right)^2=0\Leftrightarrow a=b\)