Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\sum\frac{a}{\sqrt{\left(2a\right)^2+\left(b+c\right)^2}}\le\sqrt{2}\sum\frac{a}{2a+b+c}=\sqrt{2}\sum a\left(\frac{1}{a+b+a+c}\right)\le\frac{\sqrt{2}}{4}\sum\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{3\sqrt{2}}{4}\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta có:\(\left(a^2+bc\right)\left(b+c\right)=b\left(a^2+c^2\right)+c\left(a^2+b^2\right)\)
\(\Rightarrow\sqrt{\frac{\left(a^2+bc\right)\left(b+c\right)}{a\left(b^2+c^2\right)}}=\sqrt{\frac{b\left(a^2+c^2\right)+c\left(a^2+b^2\right)}{a\left(b^2+c^2\right)}}\)
Tương tự\(\Rightarrow\)VT=\(\Sigma\sqrt{\frac{b\left(a^2+c^2\right)+c\left(a^2+b^2\right)}{a\left(b^2+c^2\right)}}\)
Đặt \(x=a\left(b^2+c^2\right)\);\(y=b\left(a^2+c^2\right)\);\(z=c\left(b^2+a^2\right)\)
VT=\(\sqrt{\frac{x+y}{z}}+\sqrt{\frac{y+z}{x}}+\sqrt{\frac{x+z}{y}}\ge3\sqrt[6]{\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}}\ge3\sqrt{2}\)(BĐT Cô-si)
Dấu''='' xra\(\Leftrightarrow\)a=b=c
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 \(a=b=c=\frac{1}{8}\)
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 \(a=b=c=\frac{1}{4}\)
Áp dụng cosi ta có \(a.a.a.b.b\le\frac{3a^5+2b^5}{5};b.b.b.a.a\le\frac{3b^5+2a^5}{5}\)
=> \(a^5+b^5\ge a^2b^2\left(a+b\right)\)
Khi đó
\(VT\le\frac{1}{ab\sqrt{a+b}}+\frac{1}{bc\sqrt{b+c}}+\frac{1}{ac\sqrt{a+c}}\)
Áp dụng BĐT buniacoxki ta có :
\((\frac{1}{ab\sqrt{a+b}}+\frac{1}{bc\sqrt{b+c}}+\frac{1}{ac\sqrt{a+c}})^2\le\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\left(\frac{1}{b^2\left(a+b\right)}+\frac{1}{c^2\left(b+c\right)}+...\right)\)
Mà 1/a^2+1/b^2+1/c^2=1(giả thiết)
=> \(VT\le VP\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c=can(3)
\(\sqrt{ab}+\sqrt{cd}\le\sqrt{\left(a+c\right)\left(b+d\right)}\)
\(\Leftrightarrow ab+cd+2\sqrt{abcd}\le ab+bc+cd+da\)
\(\Leftrightarrow bc+da\ge2\sqrt{abcd}\)
\(\Leftrightarrow bc+da-2\sqrt{abcd}\ge0\)
\(\Leftrightarrow\left(\sqrt{bc}-\sqrt{da}\right)^2\ge0\) đúng \(\forall a,b,c,d>0\)