Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(\left(\sqrt{a};\sqrt{b}\right)=\left(x;y\right)\)
\(\left(x+1\right)\left(y+1\right)=4\Leftrightarrow3=xy+x+y\le\frac{1}{4}\left(x+y\right)^2+x+y\)
\(\Rightarrow\left(x+y\right)^2+4xy-12\ge0\)
\(\Leftrightarrow\left(x+y+6\right)\left(x+y-2\right)\ge0\)
\(\Leftrightarrow x+y-2\ge0\Rightarrow x+y\ge2\)
\(P=\frac{x^4}{y^2}+\frac{y^4}{x^2}\ge\frac{\left(x^2+y^2\right)^2}{x^2+y^2}=x^2+y^2\ge\frac{1}{2}\left(x+y\right)^2\ge\frac{1}{2}.4=2\)
\(P_{min}=2\) khi \(x=y=1\) hay \(a=b=1\)
cho hỏi là cái 3 = xy + x + y ≤ \(\dfrac{1}{4}\) \(\left(x+y\right)^2\)+ x + y là như nào vậy ??
Ta có:
\(a+b+\sqrt{2\left(a+c\right)}=a+b+\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a+c}{2}}\ge3\sqrt[3]{\frac{\left(a+b\right)\left(a+c\right)}{2}}\)
Hoàn toàn tương tự ta có:
\(\frac{1}{\left(b+c+\sqrt{2\left(b+a\right)}\right)^3}\le\frac{2}{27\left(b+c\right)\left(b+a\right)}\);
\(\frac{1}{\left(c+b+\sqrt{\left(c+b\right)}\right)^3}\le\frac{2}{27\left(c+a\right)\left(c+b\right)}\)
Cộng theo bất đẳng thức trên ta được:
\(\frac{1}{\left(a+b+\sqrt{2\left(a+c\right)}\right)^3}+\frac{1}{\left(b+c+\sqrt{2\left(b+a\right)}\right)^3}+\frac{1}{\left(c+a+\sqrt{2\left(c+b\right)}\right)^3}\)
\(\le\frac{4\left(a+b+c\right)}{27\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Do đó:
\(\frac{1}{\left(a+b+\sqrt{2\left(a+c\right)}\right)^3}+\frac{1}{\left(b+c+\sqrt{2\left(b+a\right)}\right)^3}+\frac{1}{\left(c+a+\sqrt{2\left(c+b\right)}\right)^3}\)
\(\le\frac{1}{6\left(ab+bc+ca\right)}\)
Vậy bất đẳng thức được chứng minh, bất đẳng thức xày ra khi \(a=b=c=\frac{1}{4}\)
1)
\(2a+\frac{4}{a}+\frac{16}{a+2}=\left(a+\frac{4}{a}\right)+\left[\left(a+2\right)+\frac{16}{a+2}\right]-2\ge4+8-2=10\)
Dấu "=" xảy ra khi a=2
2)
\(\hept{\begin{cases}\sqrt{a\left(1-4a\right)}=\frac{1}{2}\sqrt{4a\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4a+1-4a}{2}=\frac{1}{4}\\\sqrt{b\left(1-4b\right)}=\frac{1}{2}\sqrt{4\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4b+1-4b}{2}=\frac{1}{4}\\\sqrt{c\left(1-4c\right)}=\frac{1}{2}\sqrt{4c\left(1-4c\right)}\le\frac{1}{2}\cdot\frac{4c+1-4c}{2}=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\sqrt{a\left(1-4a\right)}+\sqrt{b\left(1-4b\right)}+\sqrt{c\left(1-4c\right)}\le\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{8}\)
Đặt \(\left(a;b;c\right)=\left(x^2;y^2;z^2\right)\Rightarrow x^2+y^2+z^2\ge1\)
\(P=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{x^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}=A+B\)
\(A=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\Rightarrow A^2=\frac{x^4}{y^2}+\frac{y^4}{z^2}+\frac{z^4}{x^2}+2\left(\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{xz^2}{y}\right)\)
Mà: \(\frac{x^4}{y^2}+\frac{x^2y}{z}+\frac{x^2y}{z}+z^2\ge4x^2\)
Tương tự và cộng lại ta có:
\(A^2+\left(x^2+y^2+z^2\right)\ge4\left(x^2+y^2+z^2\right)\Rightarrow A^2\ge3\left(x^2+y^2+z^2\right)=3\)
Xét \(B=\frac{x^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}\Rightarrow B^2=\frac{x^4}{z^2}+\frac{y^4}{x^2}+\frac{z^4}{y^2}+2\left(\frac{xy^2}{z}+\frac{yz^2}{x}+\frac{zx^2}{y}\right)\)
Có: \(\frac{x^4}{z^2}+\frac{zx^2}{y}+\frac{zx^2}{y}+y^2\ge4x^2\)
\(\Rightarrow B^2\ge3\left(x^2+y^2+z^2\right)=3\) \(\Rightarrow B\ge\sqrt{3}\)
\(\Rightarrow P\ge2\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
1. Cho a,b,c thực dương thỏa mãn: abc=1
\(b^4+c^4-bc\left(b^2+c^2\right)=\left(b^2+bc+c^2\right)\left(b-c\right)^2\)
\(\Rightarrow b^4+c^4\ge bc\left(b^2+c^2\right)\)
Tương tự\(\Rightarrow\Sigma_{cyc}\frac{a}{a+b^4+c^4}\le\Sigma_{cyc}\frac{a}{a+bc\left(b^2+c^2\right)}=\Sigma_{cyc}\frac{a}{bc\left(a^2+b^2+c^2\right)}=\frac{1}{a^2+b^2+c^2}\Sigma_{cyc}\frac{a}{bc}\)
\(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}=\frac{a^2+b^2+c^2}{abc}=a^2+b^2+c^2\)
\(\Rightarrow\frac{1}{a^2+b^2+c^2}\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)=1\)
oke rồi he
@Nub :v
Áp dụng Bunhiacopski ta dễ có:
\(\frac{a}{b^4+c^4+a}=\frac{a\left(1+1+a^3\right)}{\left(b^4+c^4+a\right)\left(1+1+a^3\right)}\le\frac{a^4+2a}{\left(a^2+b^2+c^2\right)^2}\)
Tương tự:
\(\frac{b}{a^4+c^4+b}\le\frac{b^4+2b}{\left(a^2+b^2+c^2\right)^2};\frac{c}{a^4+b^4+c}\le\frac{c^4+2c}{\left(a^2+b^2+c^2\right)^2}\)
Cộng lại:
\(A\le\frac{a^4+b^4+c^4+2a+2b+2c}{\left(a^2+b^2+c^2\right)^2}\)
Ta đi chứng minh:
\(\frac{a^4+b^4+c^4+2a+2b+2c}{\left(a^2+b^2+c^2\right)^2}\le1\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\)
Cái này luôn đúng theo Cauchy
Đẳng thức xảy ra tại a=b=c=1
