Đặt \(\left\{{}\begin{matrix}a=x\\b=2y\\c=3z\end{matrix}\right.\) ta cần chứng minh

\(x^4+2y^4+3z^4\ge6\left(\dfrac{x+2y+3z}{6}\right)^4\)

Ta có:

\(x^4+2y^4+3z^4\ge\dfrac{\left(x^2+2y^2+3z^2\right)^2}{6}\ge\dfrac{1}{6}.\dfrac{\left(x+2y+3z\right)^4}{6^2}=6\left(\dfrac{x+2y+3z}{6}\right)^4\)

dòng thứ 2 ( từ dưới lên ) sao có đc nó vậy ạ

\(B=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\sqrt{1+x^2y^2}\)

\(\ge\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\dfrac{\left(1+0,25xy\right)}{\sqrt{1,0625}}\)

\(=\dfrac{1}{\sqrt{1,0625}}\left(\dfrac{1}{x}+\dfrac{1}{y}+0,25x+0,25y\right)\)

\(=\dfrac{1}{\sqrt{1,0625}}\left(\left(\dfrac{1}{x}+4x\right)+\left(\dfrac{1}{y}+4y\right)-\dfrac{15}{4}\left(x+y\right)\right)\)

\(\ge\dfrac{1}{\sqrt{1,0625}}\left(4+4-\dfrac{15}{4}\right)=\sqrt{17}\)

bạn giải thích giùm mình dòng thứ 2 đi :) dùng bđt j v :)

Ta có:

\(\sqrt{6}+\sqrt[3]{6}< \sqrt{6+\sqrt{6+...+\sqrt{6}}}+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}}< \sqrt{6+\sqrt{6+...+\sqrt{9}}}+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{8}}}\)

\(\Leftrightarrow4< \sqrt{6}+\sqrt[3]{6}< \sqrt{6+\sqrt{6+...+\sqrt{6}}}+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}}< 3+2\)

\(\Leftrightarrow4< \sqrt{6+\sqrt{6+...+\sqrt{6}}}+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}}< 5\)

phân tích 4<\(\sqrt{6}+\sqrt[3]{6}\) làm thế nào ?

\(P=xy+x+y\le\dfrac{x^2+y^2}{2}+\sqrt{2\left(x^2+y^2\right)}\)

\(=\dfrac{2017}{2}+\sqrt{2.2017}=\dfrac{2017}{2}+\sqrt{4034}\)

Ngồi gõ cả tiếng rồi ngộ ra mới out nick :|

\(pt\left(2\right)\Leftrightarrow\sqrt{2\left(x-y\right)^2+10x-6y+12}-\sqrt{y}-\sqrt{x+2}=0\)

\(\Leftrightarrow\sqrt{2\left(x-y\right)^2+10x-6y+12}-2\sqrt{y}-\left(\sqrt{x+2}-\sqrt{y}\right)=0\)

\(\Leftrightarrow\dfrac{2\left(x-y\right)^2+10x-6y+12-4y}{\sqrt{2\left(x-y\right)^2+10x-6y+12}+2\sqrt{y}}-\dfrac{x+2-y}{\sqrt{x+2}+\sqrt{y}}=0\)

\(\Leftrightarrow\dfrac{2\left(x-y+3\right)\left(x-y+2\right)}{\sqrt{2\left(x-y\right)^2+10x-6y+12}+2\sqrt{y}}-\dfrac{x+2-y}{\sqrt{x+2}+\sqrt{y}}=0\)

\(\Leftrightarrow\left(x-y+2\right)\left(\dfrac{2\left(x-y+3\right)}{\sqrt{2\left(x-y\right)^2+10x-6y+12}+2\sqrt{y}}-\dfrac{1}{\sqrt{x+2}+\sqrt{y}}\right)=0\)

\(\Rightarrow x=y-2\). Thay vào \(pt(1)\) có:

\(pt\left(1\right)\Leftrightarrow\sqrt{y^2-8\left(y-2\right)+9}-\sqrt[3]{\left(y-2\right)y+12-6\left(y-2\right)}\le1\)

\(\Leftrightarrow\sqrt{y^2-8y+25}-\sqrt[3]{y^2-8y+24}\le1\)

\(\Leftrightarrow\left(\sqrt{y^2-8y+25}-3\right)-\left(\sqrt[3]{y^2-8y+24}-2\right)\le0\)

\(\Leftrightarrow\dfrac{y^2-8y+25-9}{\sqrt{y^2-8y+25}+3}-\dfrac{y^2-8y+24-8}{\sqrt[3]{\left(y^2-8y+24\right)^2}+4+2\sqrt[3]{y^2-8y+24}}\le0\)

\(\Leftrightarrow\dfrac{\left(y-4\right)^2}{\sqrt{y^2-8y+25}+3}-\dfrac{\left(y-4\right)^2}{\sqrt[3]{\left(y^2-8y+24\right)^2}+4+2\sqrt[3]{y^2-8y+24}}\le0\)

\(\Leftrightarrow\left(y-4\right)^2\left(\dfrac{1}{\sqrt{y^2-8y+25}+3}-\dfrac{1}{\sqrt[3]{\left(y^2-8y+24\right)^2}+4+2\sqrt[3]{y^2-8y+24}}\right)\le0\)

\(\Rightarrow y=4\Rightarrow x=y-2=4-2=2\)

Vậy \(x=2;y=4\)

tội nghiệp :))

ĐKXĐ: x \(\ge\) 1

Đặt \(\sqrt{x+2}=a;\sqrt{x-1}=b\left(a>0;b\ge0\right)\)

\(\Rightarrow ab=\sqrt{\left(x-1\right)\left(x+2\right)}=\sqrt{x^2+x-2};2x+4=2a^2\)

pt <=> 2a² = 3a + ab

<=> 2a² - 3a - ab = 0

<=> 2a² - a(b + 3) = 0 (đoạn này bạn có thể phân tích thành nhân tử để làm)

Coi đây là 1 pt bậc 2 ẩn a có \(\Delta=\left(b+3\right)^2\Rightarrow\sqrt{\Delta}=b+3\) (vì b + 3 > 0)

\(\Rightarrow a_1=\dfrac{b+3+b+3}{4};a_2=\dfrac{b+3-b-3}{4}\)

\(\Leftrightarrow a=\dfrac{b+3}{2}\) (vì a > 0 nên nghiệm a₂ không thỏa mãn)

\(\Leftrightarrow2a=b+3\)

\(\Leftrightarrow2\sqrt{x+2}=\sqrt{x-1}+3\)

\(\Leftrightarrow4\left(x+2\right)=x+8+6\sqrt{x-1}\)

\(\Leftrightarrow x=2\sqrt{x-1}\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+1=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2=0\)

\(\Leftrightarrow\sqrt{x-1}-1=0\)

\(\Leftrightarrow x=2\left(TM\right)\)

Vậy ...

Điều kiện tự làm nhé.

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x+2\right)}+3\sqrt{x+2}-2\left(x+2\right)=0\)

\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{x-1}+3-2\sqrt{x+2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}=0\left(1\right)\\\sqrt{x-1}+3-2\sqrt{x+2}=0\left(2\right)\end{matrix}\right.\)

Từ (1) \(\Rightarrow x=-2\)

Từ (2) \(\Rightarrow2\sqrt{x+2}-\sqrt{x-1}=3\)

Cái này đơn giản tự giải nha.

\(\Rightarrow x=2\)

\(\left\{{}\begin{matrix}x^2+4y^2-8xy=2\\x=2y+4xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2y\right)^2-4xy-2=0\\x-2y-4xy=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x-2y=a\\4xy=b\end{matrix}\right.\)thì ta có hệ

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-b-2=0\\a-b=0\end{matrix}\right.\)

Tới đây thì đơn giản rồi nhé

\(BDT\Leftrightarrow\dfrac{\dfrac{1}{a}+\dfrac{1}{a^2}}{1+\dfrac{1}{a}+\dfrac{1}{a^2}}+\dfrac{\dfrac{1}{b}+\dfrac{1}{b^2}}{1+\dfrac{1}{b}+\dfrac{1}{b^2}}+\dfrac{\dfrac{1}{c}+\dfrac{1}{c^2}}{1+\dfrac{1}{c}+\dfrac{1}{c^2}}\le2\)

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(n,h,t\right)\) thì ta có :

\(\Leftrightarrow\dfrac{n+n^2}{1+n+n^2}+\dfrac{h+h^2}{1+h+h^2}+\dfrac{t+t^2}{1+t+t^2}\le2\)

\(\Leftrightarrow\dfrac{1}{1+n+n^2}+\dfrac{1}{1+h+h^2}+\dfrac{1}{1+t+t^2}\ge1\)

Đặt \(n=\dfrac{yz}{x^2};h=\dfrac{xz}{y^2};t=\dfrac{xy}{z^2}\)\(\Rightarrow\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\)

Và \(\dfrac{x^4}{x^4+x^2yz+y^2z^2}+\dfrac{y^4}{y^4+xy^2z+x^2z^2}+\dfrac{z^4}{z^4+xyz^2+x^2y^2}\ge1\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x^4+y^4+z^4+x^2yz+xy^2z+xyz^2+x^2y^2+y^2z^2+z^2x^2}\)

Cần cm \(\dfrac{\left(x^2+y^2+z^2\right)^2}{x^4+y^4+z^4+x^2yz+xy^2z+xyz^2+x^2y^2+y^2z^2+z^2x^2}\ge1\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2\ge x^4+y^4+z^4+x^2yz+xy^2z+xyz^2+x^2y^2+y^2z^2+z^2x^2\)

\(\Leftrightarrow x^4+y^4+z^4+2\left(x^2y^2+y^2z^2+z^2x^2\right)\ge x^4+y^4+z^4+x^2yz+xy^2z+xyz^2+x^2y^2+y^2z^2+z^2x^2\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2\ge x^2yz+xy^2z+xyz^2\left(1\right)\)

Áp dụng BĐT AM-GM ta có:

\(x^2y^2+y^2z^2=y^2\left(x^2+z^2\right)\ge2xy^2z\)

Tương tự rồi cộng theo vế ta có \(\left(1\right)\) đúng

Khi \(a=b=c=1\)

Sửa đề\(VP\le 2\) sau đó nó chính là 1 dạng của BĐT Vasc k cần thêm j cả :">

Đặt \(B=a_1+a_2+...+a_{2016}\)

\(\Rightarrow A-B=\left(a_1^3+a_2^3+...+a_{2016}^3\right)-\left(a_1+a_2+....+a_{2016}\right)\)

\(=\left(a_1^3-a_1\right)+\left(a_2^3-a_2\right)+...+\left(a_{2016}^3-a_{2016}\right)\)

\(=\left(a_1-1\right)a_1\left(a_1+1\right)+\left(a_2-1\right)a_2\left(a_2+1\right)+...+\left(a_{2016}-1\right)a_{2016}\left(a_{2016}+1\right)⋮6\)

Mà \(B⋮6\Rightarrow A⋮6\)