Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu a : \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow\left(a-b\right)^2\ge0\)
a: =>2a^2+2b^2>=a^2+2ab+b^2
=>a^2-2ab+b^2>=0
=>(a-b)^2>=0(luôn đúng)
c: =>3a^2+3b^2+3c^2>=a^2+b^2+c^2+2ab+2bc+2ac
=>2a^2+2b^2+2c^2-2ab-2bc-2ac>=0
=>(a-b)^2+(b-c)^2+(a-c)^2>=0(luôn đúng)
\(a.\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\dfrac{3}{3-\sqrt{6}}=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{3}.\sqrt{3}}{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}=\sqrt{6}-\dfrac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{3\sqrt{2}-3\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{-3\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}=-3\) \(b.\left(2\sqrt{2}-\sqrt{3}\right)^2-2\sqrt{3}\left(\sqrt{3}-2\sqrt{2}\right)=\left(2\sqrt{2}-\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)=8-3=5\) \(c.\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{3+\sqrt{5}-3+\sqrt{5}}{9-5}:\sqrt{5}=\dfrac{2\sqrt{5}}{4}.\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}.\dfrac{1}{\sqrt{5}}=\dfrac{1}{2}\) \(d.\left(3-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3+\dfrac{\sqrt{ab}-3\sqrt{a}}{\sqrt{b}-3}\right)=\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)=9-a\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ca\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^2+b^2}{ab\left(a+b\right)^3}\ge\dfrac{2ab}{ab\left(a+b\right)^3}=\dfrac{2}{\left(a+b\right)^3}\\\dfrac{b^2+c^2}{bc\left(b+c\right)^3}\ge\dfrac{2bc}{bc\left(b+c\right)^3}=\dfrac{2}{\left(b+c\right)^3}\\\dfrac{c^2+a^2}{ca\left(c+a\right)^3}\ge\dfrac{2ca}{ca\left(c+a\right)^3}=\dfrac{2}{\left(c+a\right)^3}\end{matrix}\right.\)
\(\Rightarrow VT\ge2\left[\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\right]\)
Chứng minh rằng \(2\left[\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\right]\ge\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\ge\dfrac{9}{8}\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\left\{{}\begin{matrix}2ab\le a^2+b^2\\2bc\le b^2+c^2\\2ca\le c^2+a^2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ab\le a^2-ab+b^2\\bc\le b^2-bc+c^2\\ca\le c^2-ca+a^2\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}ab\left(a+b\right)\le\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\\bc\left(b+c\right)\le\left(b+c\right)\left(b^2-bc+c^2\right)=b^3+c^3\\ca\left(c+a\right)\le\left(c+a\right)\left(c^2-ca+a^2\right)=c^3+a^3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3ab\left(a+b\right)\le3\left(a^3+b^3\right)\\3bc\left(b+c\right)\le3\left(b^3+c^3\right)\\3ca\left(c+a\right)\le3\left(c^3+a^3\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^3+3ab\left(a+b\right)+b^3\le4\left(a^3+b^3\right)\\b^3+3bc\left(b+c\right)+c^3\le4\left(b^3+c^3\right)\\c^3+3ca\left(c+a\right)+a^3\le4\left(c^3+a^3\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^3\le4\left(a^3+b^3\right)\\\left(b+c\right)^3\le4\left(b^3+c^3\right)\\\left(c+a\right)^3\le4\left(c^3+a^3\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\left(a+b\right)^3}\ge\dfrac{1}{4\left(a^3+b^3\right)}\\\dfrac{1}{\left(b+c\right)^3}\ge\dfrac{1}{4\left(b^3+c^3\right)}\\\dfrac{1}{\left(c+a\right)^3}\ge\dfrac{1}{4\left(c^3+a^3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\ge\dfrac{1}{4}\left(\dfrac{1}{a^3+b^3}+\dfrac{1}{b^3+c^3}+\dfrac{1}{c^3+a^3}\right)\)
Chứng minh rằng \(\dfrac{1}{4}\left(\dfrac{1}{a^3+b^3}+\dfrac{1}{b^3+c^3}+\dfrac{1}{c^3+a^3}\right)\ge\dfrac{9}{8}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{1}{a^3+b^3}+\dfrac{1}{b^3+c^3}+\dfrac{1}{c^3+a^3}\ge\dfrac{9}{2\left(a^3+b^3+c^3\right)}=\dfrac{9}{2}\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{1}{a^3+b^3}+\dfrac{1}{b^3+c^3}+\dfrac{1}{c^3+a^3}\right)\ge\dfrac{9}{8}\) ( đpcm )
Vậy \(2\left[\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\right]\ge\dfrac{9}{4}\)
Mà \(VT\ge2\left[\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\right]\)
\(\Rightarrow VT\ge\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{a^2+b^2}{ab\left(a+b\right)^3}+\dfrac{b^2+c^2}{bc\left(b+c\right)^3}+\dfrac{c^2+a^2}{ca\left(c+a\right)^3}\ge\dfrac{9}{4}\) ( đpcm )
2)a)\(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
c)\(a^3+b^3-a^2b-ab^2=a^2\left(a-b\right)-b^2\left(a-b\right)=\left(a-b\right)^2\left(a+b\right)\ge0\\ \Leftrightarrow a^3+b^3\ge ab\left(a+b\right)\)
b)\(a^3+b^3\ge a^2b+ab^2\Leftrightarrow4a^3+4b^3\ge a^3+b^3+3a^b+3ab^2\\ \Leftrightarrow4\left(a^3+b^3\right)\ge\left(a+b\right)^3\Leftrightarrow\dfrac{a^3+b^3}{2}\ge\left(\dfrac{a+b}{2}\right)^3\)
Bài 1:Thêm đk a > b > 0
\(VT=a+\frac{1}{b\left(a-b\right)}=\left(a-b\right)+b+\frac{1}{b\left(a-b\right)}\)
Áp dụng BĐT Cô si cho 3 số dương ta có đpcm.
Đẳng thức xảy ra khi \(a-b=b=\frac{1}{b\left(a-b\right)}\Leftrightarrow a=2;b=1\)
Bài 2: BĐT \(\Leftrightarrow\left(a-b\right)+\left(b+1\right)+\frac{4}{\left(a-b\right)\left(b+1\right)^2}\ge4\) (Thêm 1 vào hai vế +bớt + thêm b)
\(\Leftrightarrow\left(a-b\right)+\frac{1}{2}\left(b+1\right)+\frac{1}{2}\left(b+1\right)+\frac{4}{\left(a-b\right)\left(b+1\right)^2}\ge4\) (tách \(b+1=\frac{1}{2}\left(b+1\right)+\frac{1}{2}\left(b+1\right)\))
Áp dụng BĐT Cô si cho 4 số dương ta thu được đpcm.
Đẳng thức xảy ra khi \(a-b=\frac{1}{2}\left(b+1\right)=\frac{4}{\left(a-b\right)\left(b+1\right)^2}\)
\(\Leftrightarrow a=2;b=1\) (chị giải rõ ra nha, em làm tắt thôi)
Bài 3 để sau ạ, có lẽ cần thêm đk b > 0. Khi đó a/ b > 1 tức là a > b và > 0
Dự đoán điểm rơi tại a = 1; b = 1/2
Em nghĩ ra rồi nhưng ko chắc đâu.
Bài 3: Dễ thấy b > 0 => a > b > 0
Trước tiên cần giảm bậc cái đã:D
\(2a^3+1=a^3+a^3+1\ge3\sqrt[3]{a^6.1}=3a^2\)
Đẳng thức xảy ra khi a = 1 (1)
Do vậy: \(\frac{2a^3+1}{4b\left(a-b\right)}\ge\frac{3a^2}{4ab-4b^2}\). Do a > b > 0. Chia hai vế cho b2 ta được:
\(\frac{2a^3+1}{4b\left(a-b\right)}\ge\frac{3\left(\frac{a}{b}\right)^2}{4.\frac{a}{b}-4}=\frac{3t^2}{4t-4}\) với \(t=\frac{a}{b}>1\)
Ta cần chứng minh \(\frac{3t^2}{4t-4}\ge3\Leftrightarrow\frac{t^2}{4t-4}\ge1\Leftrightarrow t^2-4t+4\ge0\Leftrightarrow\left(t-2\right)^2\ge0\) (đúng)
Đẳng thức xảy ra khi a = 2b tức là theo (1) suy ra \(b=\frac{1}{2}\)
Ta có đpcm.
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
1: =>4a^3+4b^3-a^3-3a^2b-3ab^2-b^3>=0
=>a^3-a^2b-ab^2+b^3>=0
=>(a+b)(a^2-ab+b^2)-ab(a+b)>=0
=>(a+b)(a-b)^2>=0(luôn đúng)
2: \(a^4+b^4=\dfrac{a^4}{1}+\dfrac{b^4}{1}>=\dfrac{\left(a^2+b^2\right)^2}{1}=\dfrac{1}{2}\left(\dfrac{a^2}{1}+\dfrac{b^2}{1}\right)^2\)
=>\(a^4+b^4>=\dfrac{1}{2}\left(\dfrac{\left(a+b\right)^2}{2}\right)^2=\dfrac{\left(a+b\right)^4}{8}\)