\(A\ge\frac{1}{3}\left(x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}\right)^2\ge\frac{1}{3}\left(x+y+z+\frac{9}{x+y+z}\right)^2=\frac{100}{3}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)

M= \(x^2y^2+2+\frac{1}{x^2y^2}=\left(xy+\frac{1}{xy}\right)^2\)

\(xy+\frac{1}{xy}=xy+\frac{1}{16xy}+\frac{15}{16xy}\ge2\sqrt{xy.\frac{1}{16xy}}+\frac{15\left(x+y\right)}{16xy}=\frac{1}{2}+\frac{15}{16}\left(\frac{1}{x}+\frac{1}{y}\right)\ge\)\(\frac{1}{2}+\frac{15}{16}.\frac{4}{x+y}=\frac{1}{2}+\frac{15}{16}.4=\frac{17}{4}\) => M\(\ge\frac{17^2}{4^2}\)

dấu '=' khi xy = \(\frac{1}{16xy};x=y=>x=y=\frac{1}{2}\)

\(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=\frac{x^2y^2+1}{y^2}.\frac{y^2x^2+1}{x^2}=\frac{\left(x^2y^2+1\right)^2}{x^2y^2}\)

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

ta có:\(xy+\frac{1}{xy}=16xy+\frac{1}{xy}-15xy \left(1\right) \)

mặt khác:\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow x^2+y^2+2xy\ge4xy\)

\(\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\Rightarrow-15xy\ge-\frac{15}{4} \left(2\right)\)

áp dụng bất đẳng thức cô si ta có:\(16xy+\frac{1}{xy}\ge2\sqrt{16xy.\frac{1}{xy}}=8 \left(3\right)\)

từ (1), (2), (3) ta có\(xy+\frac{1}{xy}\ge8-\frac{15}{4}=\frac{17}{4}\Rightarrow\left(xy+\frac{1}{xy}\right)^2\ge\frac{289}{16}\)

vậy \(M_{min}=\frac{289}{16}\)đạt được khi \(x=y=\frac{1}{2}\)

\(\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=2+x^2y^2+\frac{1}{x^2y^2}\)

Áp dụng bđt AM-GM và bđt Cauchy-Schwarz:

\(x^2y^2+\frac{1}{x^2y^2}=x^2y^2+\frac{1}{16x^2y^2}+\frac{15}{16x^2y^2}\)

\(\ge2\sqrt{x^2y^2.\frac{1}{16x^2y^2}}+\frac{15}{16x^2y^2}=8+\frac{15}{16x^2y^2}\)

Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\Rightarrow x^2y^2\le\frac{1}{16}\Rightarrow16x^2y^2\le1\Rightarrow\frac{15}{16x^2y^2}\ge15\)

\(\Rightarrow8+\frac{15}{16x^2y^2}\ge23\)

\("="\Leftrightarrow x=y=\frac{1}{2}\)

bạn thay x=1-y rồi thay vào H sau đó làm bình thường nhé

Nguyễn Thị Diễm Quỳnhtran nguyen bao quanHoàng Đình BảoYHoàng Tử Hànguyen thi thanh huyenNgô Thành ChungHUYNH NHAT TUONG VY?Amanda?Ribi Nkok NgokLuân ĐàoPhùng Tuệ MinhTrần Trung NguyênPhạm Hoàng Hải AnhVõ Thị Tuyết KhaNguyễn Phương TrâmNguyễn Huy TúAkai HarumaLightning FarronNguyễn Thanh HằngMysterious Personsoyeon_Tiểubàng giảiVõ Đông Anh TuấnPhương AnTrần Việt Linh

\(I=2+x+\frac{1}{x}+y+\frac{1}{y}+\frac{x}{y}+\frac{y}{x}\)

\(I=2+x+\frac{1}{2x}+y+\frac{1}{2y}+\frac{x}{y}+\frac{y}{x}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(I\ge2+2\sqrt{\frac{x}{2x}}+2\sqrt{\frac{y}{2y}}+2\sqrt{\frac{xy}{xy}}+\frac{1}{2}.\frac{4}{\left(x+y\right)}\)

\(I\ge4+2\sqrt{2}+\frac{2}{x+y}\ge4+2\sqrt{2}+\frac{2}{\sqrt{2\left(x^2+y^2\right)}}=4+3\sqrt{2}\)

\(\Rightarrow I_{min}=4+3\sqrt{2}\) khi \(x=y=\frac{1}{\sqrt{2}}\)

\(K=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

Ta có: \(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\)

\(\Rightarrow\left(x+\frac{1}{x}\right)^2\ge4\)

\(\Rightarrow\left(y+\frac{1}{y}\right)^2\ge4\)

\(\Rightarrow M\ge8\)

\(K\ge\frac{1}{2}\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2=\frac{1}{2}\left(4x+\frac{1}{x}+4y+\frac{1}{y}-3\left(x+y\right)\right)^2\)

\(K\ge\frac{1}{2}\left(2\sqrt{\frac{4x}{x}}+2\sqrt{\frac{4y}{y}}-3.1\right)^2=\frac{25}{2}\)

\(\Rightarrow K_{min}=\frac{25}{2}\) khi \(x=y=\frac{1}{2}\)

\(1,\) Áp dụng BĐT: \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\text{ và }\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Dấu \("="\Leftrightarrow x=y\)

\(A=\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\\ A\ge\dfrac{1}{2}\left(1+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(1+\dfrac{4}{a+b}\right)^2+17=\dfrac{25}{2}+17=\dfrac{59}{2}\\ \text{Dấu }"="\Leftrightarrow\left\{{}\begin{matrix}a+\dfrac{1}{a}=b+\dfrac{1}{b}\\a+b=1\end{matrix}\right.\Leftrightarrow a=b=\dfrac{1}{2}\)

\(2,\text{Đặt }A=\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(\dfrac{xy^2z}{xz}+\dfrac{xyz^2}{xy}+\dfrac{x^2yz}{yz}\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(x^2+y^2+z^2\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+6\)

Áp dụng Cosi: \(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}\ge2y^2\)

CMTT: \(\left\{{}\begin{matrix}\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\ge2z^2\\\dfrac{x^2y^2}{z^2}+\dfrac{x^2z^2}{y^2}\ge2x^2\end{matrix}\right.\)

Cộng VTV \(\Leftrightarrow A^2\ge2\left(x^2+y^2+z^2\right)+6=12\\ \Leftrightarrow A\ge2\sqrt{3}\)

Dấu \("="\Leftrightarrow x=y=z=1\)

?Amanda?, Phạm Lan Hương, Phạm Thị Diệu Huyền, Vũ Minh Tuấn, Nguyễn Ngọc Lộc , @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @Trần Thanh Phương

giúp e với ạ! Cần trước 5h chiều nay! Cảm ơn mn nhiều!

Tranh thủ làm 1, 2 bài rồi ăn cơm:

1/ Đặt \(m=n-2008>0\)

\(\Rightarrow2^{2008}\left(369+2^m\right)\) là số chính phương

\(\Rightarrow369+2^m\) là số chính phương

m lẻ thì số trên chia 3 dư 2 nên ko là số chính phương

\(\Rightarrow m=2k\Rightarrow369=x^2-\left(2^k\right)^2=\left(x-2^k\right)\left(x+2^k\right)\)

b/

\(2\left(a^2+b^2\right)\left(a+b-2\right)=a^4+b^4\) \(\left(a+b>2\right)\)

\(\Rightarrow2\left(a^2+b^2\right)\left(a+b-2\right)\ge\frac{1}{2}\left(a^2+b^2\right)^2\)

\(\Rightarrow a^2+b^2\le4\left(a+b-2\right)\)

\(\Rightarrow\left(a-2\right)^2+\left(b-2\right)^2\le0\Rightarrow a=b=2\)

\(\Rightarrow x=y=4\)

2/

\(A\ge\frac{8a^2+1-a}{4a}+b^2=2a+\frac{1}{4a}+b^2-\frac{1}{4}=a+\frac{1}{4a}+b^2+a-\frac{1}{4}\)

\(A\ge a+\frac{1}{4a}+b^2+1-b-\frac{1}{4}=a+\frac{1}{4a}+\left(b-\frac{1}{2}\right)^2+\frac{1}{2}\ge1+\frac{1}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)

b/ Giả thiết tương đương:

\(a\left(a+1\right)+b\left(b+1\right)=2\left(a+1\right)\left(b+1\right)\)

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

Hình như bạn ghi nhầm biểu thức

Đặt \(\left(\frac{a}{b+1};\frac{b}{a+1}\right)=\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}x+y=2\\0\le x;y\le2\end{matrix}\right.\)

\(P=\left(1+x^3\right)\left(1+y^3\right)=1+x^3+y^3+\left(xy\right)^3\)

\(=1+\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3\)

\(=\left(xy\right)^3-6xy+9=9-xy\left(6-\left(xy\right)^2\right)\)

Do \(xy\le1\Rightarrow6-\left(xy\right)^2>0\Rightarrow xy\left(6-\left(xy\right)^2\right)\ge0\)

\(\Rightarrow P\le9\Rightarrow P_{max}=9\) khi \(\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\) hay \(\left(a;b\right)=\left(0;2\right);\left(2;0\right)\)

Câu c giống câu này:

https://hoc24.vn/hoi-dap/question/790896.html

Bạn tham khảo tạm, cách đó quá dài nên chắc chắn ko tối ưu, nó trâu bò quá