chứng minh
\(\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a}\ge4\), \(\left(a,b,c,d>0\right)\)
Chứng minh các bất đẳng thức sau:
1. \(\frac{3}{a+b}+\frac{2}{c+d}+\frac{a+b}{\left(a+c\right)\left(b+d\right)}\ge\frac{12}{a+b+c+d}\)
2. \(\frac{\left(a+b\right)^2}{a+b-c}+\frac{\left(b+c\right)^2}{-a+b+c}+\frac{\left(c+a\right)^2}{a-b+c}\ge4.\left(a+b+c\right)\)
chứng minh các BĐT
1.\(\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{b+d}{d+a}\ge4\)với a,b,c,d >0
2.\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge4\left(\frac{1}{2a+b+c}+\frac{1}{2b+c+d}+\frac{1}{2c+d+a}+\frac{1}{2d+a+b}\right)\)
3.\(\frac{1}{a^4+b^4+c^4}+\frac{2}{a^2b^2+b^2c^2+c^2a^2}\ge\left(\frac{3}{a^2+b^2+c^2}\right)^2\\ \)với a,b,c>0
4.\(\frac{1}{3x-2}-\frac{1}{x-10}+\frac{1}{13-2x}\ge\frac{3}{7}\)vói x,y t/m\(\frac{2}{3}< x< \frac{13}{2}\)
Bài 1:Cho a,b,c,d là các số dương. Chứng minh rằng :
\(\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}+\frac{b^4}{\left(b+c\right)\left(b^2+c^2\right)}+\frac{c^4}{\left(c+d\right)\left(c^2+d^2\right)}+\frac{d^4}{\left(d+a\right)\left(d^2+a^2\right)}\ge\frac{a+b+c+d}{4}\)
Bài 2:Cho \(a>0,b>0,c>0\).\(CM:\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Bài 3: a) Cho x,y,>0. CMR:\(\frac{x^3}{x^2+xy+y^2}\ge\frac{2x-y}{3}\)
b) Chứng minh rằng\(\Sigma\frac{a^3}{a^2+ab+b^2}\ge\frac{a+b+c}{3}\)
Xét \(\frac{a^3}{a^2+ab+b^2}-\frac{b^3}{a^2+ab+b^2}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=a-b\)
Tương tự, ta được: \(\frac{b^3}{b^2+bc+c^2}-\frac{c^3}{b^2+bc+c^2}=b-c\); \(\frac{c^3}{c^2+ca+a^2}-\frac{a^3}{c^2+ca+a^2}=c-a\)
Cộng theo vế của 3 đẳng thức trên, ta được: \(\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\)\(-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)=0\)
\(\Rightarrow\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\)\(=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\)
Ta đi chứng minh BĐT phụ sau: \(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)(*)
Thật vậy: (*)\(\Leftrightarrow\frac{2}{3}\left(a-b\right)^2\ge0\)*đúng*
\(\Rightarrow2LHS=\Sigma_{cyc}\frac{a^3+b^3}{a^2+ab+b^2}=\Sigma_{cyc}\text{ }\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}\)\(\ge\Sigma_{cyc}\text{ }\frac{\frac{1}{3}\left(a+b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=\frac{1}{3}\text{}\Sigma_{cyc}\left[\left(a+b\right)\right]=\frac{2\left(a+b+c\right)}{3}\)
\(\Rightarrow LHS\ge\frac{a+b+c}{3}=RHS\)(Q.E.D)
Đẳng thức xảy ra khi a = b = c
P/S: Có thể dùng BĐT phụ ở câu 3a để chứng minhxD:
1) ta chứng minh được \(\Sigma\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}=\Sigma\frac{b^4}{\left(a+b\right)\left(a^2+b^2\right)}\)
\(VT=\frac{1}{2}\Sigma\frac{a^4+b^4}{\left(a+b\right)\left(a^2+b^2\right)}\ge\frac{1}{4}\Sigma\frac{a^2+b^2}{a+b}\ge\frac{1}{8}\Sigma\left(a+b\right)=\frac{a+b+c+d}{4}\)
bài 2 xem có ghi nhầm ko
3a biến đổi tí là xong
b tuong tự bài 1
1/ Biết \(\frac{a}{b}=\frac{c}{d}\), chứng minh
a) \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) \(\left(\frac{a-d}{c-b}\right)^4=\frac{a^4+b^4}{c^4+d^4}\)
2/ Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Chứng minh \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{b}\)
3/ Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Chứng minh a=b=c
Mình chỉ làm bài 1a, và bài 3 thôi nhé,còn lại là bạn tự làm nhé
Bài 1:
a, Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(\Rightarrow\left[\frac{a}{b}\right]^2=\left[\frac{c}{d}\right]^2=\left[\frac{a+c}{b+d}\right]^2\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{(a+c)^2}{(b+d)^2}\Rightarrow\frac{a^2+c^2}{b^2+d^2}=\frac{(a+c)^2}{(b+d)^2}\)
Bài 3 : Sửa đề : Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)
CM : a = b = c
Cách 1 : Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)
vì \(a+b+c\ne0\)
\(\frac{a}{b}=1\Rightarrow a=b;\frac{b}{c}=1\Rightarrow b=c\)
Do đó : \(a=b=c\).
Cách 2 : Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=m\), ta có : \(a=bm,b=cm,c=am\)
Do đó : \(a=bm=m(mc)=m\left[m(ma)\right]\)
\(\Rightarrow a=m^3a\Rightarrow m^3=1(a\ne0)\Rightarrow m=1\)
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\Rightarrow a=b=c\)
Cách 3 : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\Rightarrow\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}=\left[\frac{a}{b}\right]^3\Rightarrow1=\left[\frac{a}{b}\right]^3\Rightarrow\frac{a}{b}=1\)
Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\Rightarrow a=b=c\)
Cho \(\frac{a}{b}=\frac{c}{d}.\) Và b+c+d khác 0. Chứng minh rằng \(\frac{a}{d}=\frac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\)
Đề phải thêm là \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\) nhé.
Ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}.\)
\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a+b+c}{b+c+d}.\frac{a+b+c}{b+c+d}.\frac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}\)
\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}.\)
\(\Rightarrow\frac{a}{d}=\frac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\left(đpcm\right).\)
Chúc bạn học tốt!
Nhanh giùm với ạ TTTT
Cho a , b , c , d là các số thực không âm thỏa mãn
\(\left(a+b+c\right)\left(b+c+d\right)\left(c+d+a\right)\left(d+a+b\right)>0\)
Chứng minh rằng \(\sqrt{\frac{a}{b+c+d}}+\sqrt{\frac{b}{c+d+a}}+\sqrt{\frac{c}{d+a+b}}+\sqrt{\frac{d}{a+b+c}}\ge2\)
Xét: \(\sqrt{\frac{a}{b+c+d}}=\frac{\sqrt{a}}{\sqrt{b+c+d}}=\frac{a}{\sqrt{a\left(b+c+d\right)}}\)
\(\sqrt{\frac{b}{c+d+a}}=\frac{\sqrt{b}}{\sqrt{c+d+a}}=\frac{b}{\sqrt{b\left(c+d+a\right)}}\)
\(\sqrt{\frac{c}{d+a+b}}=\frac{\sqrt{c}}{\sqrt{d+a+b}}=\frac{c}{\sqrt{c\left(d+a+b\right)}}\)
\(\sqrt{\frac{d}{a+b+c}}=\frac{\sqrt{d}}{\sqrt{a+b+c}}=\frac{d}{\sqrt{d\left(a+b+c\right)}}\)
\(\Rightarrow VT=\frac{a}{\sqrt{a\left(b+c+d\right)}}+\frac{b}{\sqrt{b\left(c+d+a\right)}}+\frac{c}{\sqrt{c\left(d+a+b\right)}}+\frac{d}{\sqrt{d\left(a+b+c\right)}}\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{\begin{matrix}\sqrt{a\left(b+c+d\right)}\le\frac{a+b+c+d}{2}\\\sqrt{b\left(c+d+a\right)}\le\frac{a+b+c+d}{2}\\\sqrt{c\left(d+a+b\right)}\le\frac{a+b+c+d}{2}\\\sqrt{d\left(a+b+c\right)}\le\frac{a+b+c+d}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{a}{\sqrt{a\left(b+c+d\right)}}\ge\frac{2a}{a+b+c+d}\\\frac{b}{\sqrt{b\left(c+d+a\right)}}\ge\frac{2b}{a+b+c+d}\\\frac{c}{\sqrt{c\left(d+a+b\right)}}\ge\frac{2c}{a+b+c+d}\\\frac{d}{\sqrt{d\left(a+b+c\right)}}\ge\frac{2d}{a+b+c+d}\end{matrix}\right.\)
\(\Rightarrow VT\ge\frac{2a}{a+b+c+d}+\frac{2b}{a+b+c+d}+\frac{2c}{a+b+c+d}+\frac{2d}{a+b+c+d}\)
\(\Rightarrow VT\ge\frac{2\left(a+b+c+d\right)}{a+b+c+d}\)
\(\Rightarrow VT\ge2\)
\(\Rightarrow\frac{a}{\sqrt{a\left(b+c+d\right)}}+\frac{b}{\sqrt{b\left(c+d+a\right)}}+\frac{c}{\sqrt{c\left(d+a+b\right)}}+\frac{d}{\sqrt{d\left(a+b+c\right)}}\ge2\)
\(\Leftrightarrow\sqrt{\frac{a}{b+c+d}}+\sqrt{\frac{b}{c+d+a}}+\sqrt{\frac{c}{d+a+b}}+\sqrt{\frac{d}{a+b+c}}\ge2\) ( đpcm )
Lời giải:
Áp dụng bất đẳng thức AM-GM:
\(\frac{b+c+d}{a}=\frac{b+c+d}{a}.1\leq \left(\frac{\frac{b+c+d}{a}+1}{2}\right)^2=\left(\frac{b+c+d+a}{2a}\right)^2\)
\(\sqrt{\frac{a}{b+c+d}}\geq \frac{2a}{a+b+c+d}\). Tương tự với các phân thức còn lại:
\(\Rightarrow \text{VT}\geq \frac{2(a+b+c+d)}{a+b+c+d}=2\) (đpcm)
\(\frac{a}{b}=\frac{c}{d}\)(a,b,c,d khác 0)
Chứng minh :\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Đặt: \(\frac{a}{b}=\frac{c}{d}=k\)
==> a = b.k
c = d.k
Ta có : \(\frac{a^2+b^2}{c^2+d^2}\) = \(\frac{b^2.k^2+b^2}{d^2.k^2+d^2}\) = \(\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}\) = \(\frac{b^2}{d^2}\) (1)
\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) = \(\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\) = \(\frac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}\) = \(\frac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}\) = \(\frac{b^2}{d^2}\) (2)
Từ (1) và (2) ==> \(\frac{a^2+b^2}{c^2+d^2}\) = \(\frac{\left(a-b\right)^2}{\left(c-d^{ }\right)^2}\) (đpcm)
Good for you
Cho \(\frac{a}{b}=\frac{c}{d}\)(b,d ≠ 0; b≠ d). Chứng minh rằng : \(\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có
\(VT:\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{b^{2018}\cdot k^{2018}+d^{2018}\cdot k^{2018}}{b^{2018}+d^{2018}}=\frac{k^{2018}\left(b^{2018}+d^{2018}\right)}{b^{2018}+d^{2018}}=k^{2018}\)
\(VP:\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{\left(bk+dk\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{k^{2018}\cdot\left(b+d\right)^{2018}}{\left(b+d\right)^{2018}}=k^{2018}\)
\(\Rightarrow VT=VP\)
Hay \(\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\left(đpcm\right)\)
1.Chứng minh rằng :
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+b+c+d\)với \(a\ge-1;b\ge-4;c\ge2;d>3\)
2. Chứng minh rằng :
\(\frac{a^2}{b^5}+\frac{b^2}{c^5}+\frac{c^2}{d^5}+\frac{d^2}{a^5}\ge\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)với \(a,b,c,d>0\)
Câu 1:
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+1+b+4+c-2+d-3=a+b+c+d\)
Dấu = xảy ra khi a = -1; b = -4; c = 2; d= 3
\(\frac{a^2}{b^5}+\frac{1}{a^2b}\ge\frac{2}{b^3}\)\(\Leftrightarrow\)\(\frac{a^2}{b^5}\ge\frac{2}{b^3}-\frac{1}{a^2b}\)
\(\frac{2}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\)\(\Leftrightarrow\)\(\frac{1}{a^2b}\le\frac{2}{3a^3}+\frac{1}{3b^3}\)
\(\Rightarrow\)\(\Sigma\frac{a^2}{b^5}\ge\Sigma\left(\frac{5}{3b^3}-\frac{2}{3a^3}\right)=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)
ta sẽ giết ngươi kí tên dép đờ kiu lờ