Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1
\(VT=\dfrac{a^2}{ab^2+abc+ac^2}+\dfrac{b^2}{c^2b+abc+a^2b}+\dfrac{c^2}{a^2c+abc+b^2c}\)
Áp dụng bđt Cauchy dạng phân thức
\(\Rightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b\right)+abc+ac\left(a+c\right)+abc+bc\left(b+c\right)+abc}\)
\(\Leftrightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}\)
\(\Leftrightarrow VT\ge\dfrac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)
Dấu ''='' xảy ra khi \(a=b=c\)
Bài 2
\(VT=\left(\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}\right)\left[\left(\dfrac{\sqrt{a}}{b+c}\right)^2+\left(\dfrac{\sqrt{b}}{c+a}\right)^2+\left(\dfrac{\sqrt{c}}{a+b}\right)^2\right]\)
Áp dụng bđt Bunhiacopxki ta có
\(VT\ge\left(\sqrt{a}.\dfrac{\sqrt{a}}{b+c}+\sqrt{b}.\dfrac{\sqrt{b}}{c+a}+\sqrt{c}.\dfrac{\sqrt{c}}{a+b}\right)^2\)
\(\Leftrightarrow VT\ge\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\)
Xét \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Áp dụng bđt Cauchy dạng phân thức ta có
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ac\right)}=\dfrac{3}{2}\)
\(\Rightarrow\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\ge\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)
\(\Rightarrow VT\ge\dfrac{9}{4}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c\)
Đặt \(A=x+\dfrac{4}{\left(x-y\right)\left(y+1\right)^2}\ge3\)
\(=\left(x-y\right)+\dfrac{4}{\left(x-y\right)\left(y+1\right)^2}+\left(y+1\right)-1\)
Áp dụng BĐT Cô-si cho 2 số dương ta có :
\(\left(x-y\right)+\dfrac{4}{\left(x-y\right)\left(y+1\right)^2}\ge2\sqrt{\dfrac{\left(x-y\right).4}{\left(x-y\right)\left(y+1\right)^2}}=\dfrac{4}{y+1}\)
Xảy ra khi : \(\left(x-y\right)\left(y+1\right)=2\) ( do \(a,b>0\))
\(\Rightarrow A\ge\dfrac{4}{y+1}+\left(y+1\right)-1\)
Sử dụng Cô-Si lần nữa, ta có :
\(\dfrac{4}{y+1}+\left(y+1\right)\ge2\sqrt{\dfrac{4}{y+1}\left(y+1\right)}=2.2=4\)
Xảy ra khi \(\left(y+1\right)^2=4\Leftrightarrow y=1\)
Từ đây ta có thể thấy : \(A\ge4-1=3\)
Dấu "=" xảy ra khi \(\left(x-y\right)\cdot\left(y+1\right)=2\) và \(y=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right..\)
Bài này hồi lúc cũng không biết làm, h biết truyền lại cho bạn :D
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
d/ \(\Leftrightarrow a^4-a^3b+b^4-ab^3\ge0\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^3-b^3\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\) (luôn đúng)
e/ \(\Leftrightarrow a^6+b^6+a^5b+ab^5\ge a^6+b^5+a^4b^2+a^2b^4\)
\(\Leftrightarrow a^5b-a^4b^2+ab^5-a^2b^4\ge0\)
\(\Leftrightarrow a^4b\left(a-b\right)-ab^4\left(a-b\right)\ge0\)
\(\Leftrightarrow ab\left(a-b\right)\left(a^3-b^3\right)\ge0\)
\(\Leftrightarrow ab\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\) (luôn đúng)
f/ \(\frac{a^6}{b^2}+a^2b^2\ge2\sqrt{\frac{a^8b^2}{b^2}}=2a^4\) ; \(\frac{b^6}{a^2}+a^2b^2\ge2b^4\)
\(\Rightarrow\frac{a^6}{b^2}+\frac{b^6}{a^2}\ge2a^4+2b^4-2a^2b^2\)
\(\Leftrightarrow\frac{a^6}{b^2}+\frac{b^6}{a^2}\ge a^4+b^4+\left(a^4+b^4-2a^2b^2\right)\)
\(\Leftrightarrow\frac{a^6}{b^2}+\frac{b^6}{a^2}\ge a^4+b^4+\left(a^2-b^2\right)^2\ge a^4+b^4\)
a/ \(VT=a^2\left(1+b^2\right)+b^2\left(1+c^2\right)+c^2\left(1+a^2\right)\)
\(VT=a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2\)
\(VT\ge6\sqrt[6]{a^6b^6c^6}=6\left|abc\right|\ge6abc\)
Dấu "=" xảy ra khi \(a=b=c=1\)
b/ \(\Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2\ge4ab+4ac+4ad+4ae\)
\(\Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+\left(a-2e\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(\frac{a}{2}=b=c=d=e\)
c/ \(\Leftrightarrow\frac{a^3+b^3}{2}\ge\frac{a^3+b^3+3a^2b+3ab^2}{8}\)
\(\Leftrightarrow a^3-a^2b+b^3-ab^2\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(a=b\)
Lời giải:
Không mất tính tổng quát, giả sử \(a>b> c\). Khi đó \(a-b>0; b-c> 0; c-a< 0\)
Áp dụng BĐT AM-GM:\(\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}\geq \frac{2}{(a-b)(b-c)}\)
Tiếp tục áp dụng BĐT AM-GM: \((a-b)(b-c)\leq \left(\frac{a-b+b-c}{2}\right)^2=\frac{(c-a)^2}{4}\)
\(\Rightarrow \frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}\geq \frac{2}{\frac{(c-a)^2}{4}}=\frac{8}{(c-a)^2}\)
\(\Rightarrow \frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\geq \frac{9}{(c-a)^2} \)
Mà \(0\leq c< a\leq 2\Rightarrow 0< a-c\leq 2\Rightarrow (c-a)^2=(a-c)^2\leq 4\)
\(\Rightarrow \frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\geq \frac{9}{(c-a)^2} \geq \frac{9}{4}\) (đpcm)
Dấu "=" xảy ra khi $(a,b,c)=(2,1,0)$ và hoán vị.
Lời giải:
Áp dụng BĐT Cauchy ta có:
\(\frac{a^4}{b^3(c+a)}+\frac{c+a}{4a}+\frac{1}{2}\geq 3\sqrt[3]{\frac{a^3}{8b^3}}=\frac{3a}{2b}\)
\(\frac{b^4}{c^3(a+b)}+\frac{a+b}{4b}+\frac{1}{2}\geq 3\sqrt[3]{\frac{b^3}{8c^3}}=\frac{3b}{2c}\)
\(\frac{c^4}{a^3(b+c)}+\frac{b+c}{4c}+\frac{1}{2}\geq 3\sqrt[3]{\frac{c^3}{8a^3}}=\frac{3c}{2a}\)
Cộng theo vế và rút gọn:
\(\Rightarrow \frac{a^4}{b^3(c+a)}+\frac{b^4}{c^3(a+b)}+\frac{c^4}{a^3(b+c)}+\frac{1}{4}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+\frac{9}{4}\geq \frac{3}{2}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\)
\(\Rightarrow \frac{a^4}{b^3(c+a)}+\frac{b^4}{c^3(a+b)}+\frac{c^4}{a^3(b+c)}\geq \frac{5}{4}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})-\frac{9}{4}\geq \frac{5}{4}.3\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}-\frac{9}{4}\)
hay \( \frac{a^4}{b^3(c+a)}+\frac{b^4}{c^3(a+b)}+\frac{c^4}{a^3(b+c)}\geq \frac{5}{4}.3-\frac{9}{4}=\frac{3}{2}\)
Ta có đpcm
Dấu bằng xảy ra khi \(a=b=c\)
Cách khác:
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\frac{(\frac{a^2}{b})^2}{b(c+a)}+\frac{(\frac{b^2}{c})^2}{c(a+b)}+\frac{(\frac{c^2}{a})^2}{a(b+c)}\geq \frac{\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2}{b(c+a)+c(a+b)+a(b+c)}\)
Tiếp tục áp dụng BĐT Cauchy-Schwarz:
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \frac{(a+b+c)^2}{b+c+a}=a+b+c\)
\(\Rightarrow \left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2\geq (a+b+c)^2\)
Do đó: \(\text{VT}\geq \frac{(a+b+c)^2}{2(ab+bc+ac)}\)
Theo hệ quả quen thuộc của BĐT Cauchy: \((a+b+c)^2\geq 3(ab+bc+ac)\)
Suy ra: \(\text{VT}\geq \frac{3(ab+bc+ac)}{2(ab+bc+ac)}=\frac{3}{2}\) (đpcm)
Áp dung BĐT schur với k=2 ta được:
a2(a-b)(a-c)+b2(b-c)(b-a)+c2(c-a)(c-b)\(\ge\)0
a4+b4+c4+abc(a+b+c)\(\ge\)ab(a2+b2)+bc(b2+c2)+ca(c2+a2)
VL Vấn đề của tui là chưa chứng minh được schur bạn ạ ~ Chứng minh giúp schur giúp
Bài 1:
Ta có: \(\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}=\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\)
Áp dụng bđt Cauchy Schwarz có:
\(\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+8bc}+b\sqrt{b^2+8bc}+c\sqrt{c^2+8bc}}\)
Lại sử dụng bđt Cauchy schwarz ta có:
\(a\sqrt{a^2+8bc}+b\sqrt{b^2+8ac}+c\sqrt{c^2+8ab}=\sqrt{a}\cdot\sqrt{a^3+8abc}+\sqrt{b}\cdot\sqrt{b^3+8abc}+\sqrt{c}\cdot\sqrt{c^3+8abc}\ge\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}}=\sqrt{\dfrac{\left(a+b+c\right)^3}{a^3+b^3+c^3+24abc}}\)
=> Ta cần chứng minh: \(\left(a+b+c\right)^3\ge a^3+b^3+c^3+24abc\)
hay \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
Áp dụng bđt Cosi ta có:
\(a+b\ge2\sqrt{ab};b+c\ge2\sqrt{bc};c+a\ge2\sqrt{ca}\)
Nhân các vế của 3 bđt trên ta đc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}=8\sqrt{a^2b^2c^2}=8abc\)
=> Đpcm