Đề ??

\(bx^2=ay^{2^{ }}=\dfrac{x^2}{\dfrac{1}{b}}=\dfrac{y^2}{\dfrac{1}{a}}=\dfrac{x^2+y^2}{\dfrac{a+b}{ab}}=\dfrac{ab}{a+b}.\)

\(\Leftrightarrow\dfrac{x^2}{a}=\dfrac{1}{a+b}=\dfrac{y^2}{b}.\)

\(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\left(\dfrac{x^2}{a}\right)^{1008}+\left(\dfrac{y^2}{b}\right)^{1008}=2.\left(\dfrac{1}{a+b}\right)^{1008}=\dfrac{2}{\left(a +b\right)^{1008}}\left(dpcm\right)\)

Theo bài ra ta có:

\(bx^2=ay^2\) \(\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}\)

\(x^2+y^2=1\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)

\(\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{1}{a+b}\) \(\left(1\right)\)

Từ \(\left(1\right)\) suy ra :

\(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}\) \(=\dfrac{\left(x^2\right)^{1008}}{a^{1008}}+\dfrac{\left(y^2\right)^{1008}}{b^{1008}}\)

\(=\left(\dfrac{x^2}{a}\right)^{1008}+\left(\dfrac{y^2}{b}\right)^{1008}\)

\(=\left(\dfrac{1}{a+b}\right)^{1008}+\left(\dfrac{1}{a+b}\right)^{1008}\)

\(=2\cdot\left(\dfrac{1}{a+b}\right)^{1008}\)

\(=2\cdot\dfrac{1^{1008}}{\left(a+b\right)^{1008}}\)

\(=2\cdot\dfrac{1}{\left(a+b\right)^{1008}}\)

\(=\dfrac{2}{a+b}^{1008}\)

Vậy \(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\dfrac{2}{a+b}^{1008}\)

Ta có: \(\hept{\begin{cases}x^2+y^2=1\\\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\end{cases}}\)

\(\Leftrightarrow b\left(a+b\right)x^4+a\left(a+b\right)y^4=ab\left(x^4+2x^2y^2+y^4\right)\)

\(\Leftrightarrow b^2x^4+a^2y^4-2abx^2y^2=0\)

\(\Leftrightarrow\left(bx^2-ay^2\right)^2=0\)

\(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)

\(\Rightarrow\frac{x^{2016}}{a^{1008}}=\frac{y^{2016}}{b^{1008}}=\frac{1}{\left(a+b\right)^{1008}}\)

\(\Rightarrow\frac{x^{2016}}{a^{1008}}+\frac{y^{2016}}{b^{21008}}=\frac{2}{\left(a+b\right)^{1008}}\)

Em vào câu hỏi tương tự tham khảo: 

Ta có: \(x^2+y^2=1\Leftrightarrow x^4+2x^2y^2+y^4=1\)

Khi đó: \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{x^4+2x^2y^2+y^4}{a+b}\)

<=> \(\left(a+b\right)\left(\frac{x^4}{a}+\frac{y^4}{b}\right)=x^4+2x^2y^2+y^4\)

<=> \(\frac{b}{a}x^4+\frac{a}{b}y^4=2x^2y^2\)

<=> \(\frac{x^4}{a^2}+\frac{y^4}{b^2}-\frac{2x^2y^2}{ab}=0\)

<=> \(\left(\frac{x^2}{a}-\frac{y^2}{b}\right)^2=0\)

<=> \(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)( dãy tỉ số bằng nhau)

Khi đó: \(\frac{x^{2016}}{a^{1008}}+\frac{y^{2016}}{b^{1008}}=2\frac{x^{2016}}{a^{1008}}=\frac{2}{\left(a+b\right)^{1008}}\)

1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

1.

- Với \(a+b\ge4\Rightarrow A\le0\)

- Với \(a+b< 4\Rightarrow4-a-b>0\)

\(\Rightarrow A=\dfrac{a}{2}.\dfrac{a}{2}.b.\left(4-a-b\right)\)

\(\Rightarrow A\le\dfrac{1}{64}\left(\dfrac{a}{2}+\dfrac{a}{2}+b+4-a-b\right)^4=4\)

\(A_{max}=4\) khi \(\left(a;b\right)=\left(2;1\right)\)

2.

\(P=a+\dfrac{1}{2}.a.2b\left(1+2c\right)\le a+\dfrac{a}{8}\left(2b+1+2c\right)^2\)

\(P\le a+\dfrac{a}{8}\left(7-2a\right)^2=\dfrac{1}{8}\left(4a^3-28a^2+57a-36\right)+\dfrac{9}{2}\)

\(P\le\dfrac{1}{8}\left(a-4\right)\left(2a-3\right)^2+\dfrac{9}{2}\le\dfrac{9}{2}\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{3}{2};1;\dfrac{1}{2}\right)\)

Câu 3 bạn xem lại đề, mình có thể chắc chắn với bạn là đề sai

Ví dụ bạn cho \(x=98,y=100\) thì vế trái chỉ lớn hơn 8 một chút

Đề đúng phải là: \(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{16xy}{\left(x-y\right)^2}\ge12\)

Nếu câu 3 đề là \(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{16xy}{\left(x-y\right)^2}\ge12\)

Ta có:

\(VT=2+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{16xy}{\left(x-y\right)^2}=\dfrac{x^2+y^2}{xy}+\dfrac{16xy}{\left(x-y\right)^2}+2\)

\(VT=\dfrac{x^2+y^2-2xy+2xy}{xy}+\dfrac{16xy}{\left(x-y\right)^2}+2\)

\(VT=\dfrac{\left(x-y\right)^2}{xy}+\dfrac{16xy}{\left(x-y\right)^2}+4\ge2\sqrt{\dfrac{16xy\left(x-y\right)^2}{xy\left(x-y\right)^2}}+4=12\)