a) \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{5}\)

\(\Leftrightarrow\frac{2015}{a+b}+\frac{2015}{b+c}+\frac{2015}{c+a}=403\)

\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=403\)

\(\Leftrightarrow3+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=403\)

\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=400\)

b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\hept{\begin{cases}a=ck\\b=dk\end{cases}}\)

Thay vào rồi c/m nhé

a)  Từ đẳng thức : \(A=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)

\(\Rightarrow A+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)

\(\Rightarrow A+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)

\(\Rightarrow A+3=\left(a+b+c\right).\frac{1}{b+c}+\left(a+b+c\right).\frac{1}{a+c}+\left(a+b+c\right).\frac{1}{a+b}\)

\(\Rightarrow A+3=\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Rightarrow A+3=2015.\frac{1}{5}\)

\(\Rightarrow A+3=403\)

\(\Rightarrow A=400\)

Vậy A = 400

b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

Khi đó : \(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2\left(bk\right)^2-3b^2k+5b^2}{2\left(bk\right)^2+3b^2k}=\frac{2k^2b^2-3b^2k+5b^2}{2b^2k^2+3b^2k}=\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2k^2+3k\right)}\)

\(=\frac{2k^2-3k+5}{2k^2+3k}\left(1\right)\);

\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2\left(dk\right)^2-3d^2k+5d^2}{2\left(dk\right)^2+3d^2k}=\frac{2d^2k^2-3d^2k+5d^2}{2d^2k^2+3d^2k}=\frac{d^2.\left(2k^2-3k+5\right)}{d^2\left(2k^2+3k\right)}\)

\(=\frac{2k^2-3k+5}{2k^2+3k}\left(2\right)\)

Từ (1) và (2) => \(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2c^2-3cd+5d^2}{2c^2+3cd}\)(đpcm)

lớp 6 mà

lớp 9 đó

\(a)\) Ta có : 

\(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{1}{2015}\)

\(\Leftrightarrow\)\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\left(a+b+c\right).\frac{1}{2015}\)

\(\Leftrightarrow\)\(\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=\frac{a+b+c}{2015}\)

\(\Leftrightarrow\)\(1+\frac{c}{a+b}+1+\frac{b}{a+c}+1+\frac{a}{b+c}=\frac{2015}{2015}\)

\(\Leftrightarrow\)\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1-3\)

\(\Leftrightarrow\)\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=-2\)

Vậy ...

Ta có : \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2015.5\)

\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a+b}{a+b}+\frac{a+c}{c+a}+\frac{b+c}{b+c}=2015.5\)

\(\Leftrightarrow Q+3=2015.5\Rightarrow Q=2015.5-3=10072\)

\(\frac{2015}{a+b}+\frac{2015}{b+c}+\frac{2015}{c+a}=\frac{2015}{90}\)

\(\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2015}{90}\)

\(1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2015}{90}\)

\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{b+a}=\frac{2015}{90}-3=\frac{349}{18}\)

Bài này cần chú ý: \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{\left(a-b\right)^2}{ab}+\frac{\left(a-c\right)\left(b-c\right)}{ac}\)

Và \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2}=\frac{\left(a-b\right)^2}{\left(a+c\right)\left(b+c\right)}+\frac{\left(a+b+2c\right)\left(a-c\right)\left(b-c\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

Thêm 3 vào 2 vế ta cần chứng minh:

\(\frac{2}{1-a}+\frac{2}{1-b}+\frac{2}{1-c}\le2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3}{2}\right)\)

\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\le\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3}{2}\) (chia hai vế cho 2 và chú ý 1 =a + b + c)

\(\Leftrightarrow\frac{3}{2}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\le\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)

\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2}\le\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\)

\(\Leftrightarrow\frac{\left(a-b\right)^2}{\left(a+c\right)\left(b+c\right)}+\frac{\left(a+b+2c\right)\left(a-c\right)\left(b-c\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\frac{\left(a-b\right)^2}{ab}+\frac{\left(a-c\right)\left(b-c\right)}{ac}\)

\(\Leftrightarrow\left(a-b\right)^2\left(\frac{1}{ab}-\frac{1}{\left(a+c\right)\left(b+c\right)}\right)+\left(\frac{1}{ac}-\frac{a+b+2c}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\right)\left(a-c\right)\left(b-c\right)\ge0\)

Quy đồng mỗi cái ngoặc to phía sau là thấy nó > 0:D

Giả sử c = min{a,b,c} như vậy (a-c)(b-c)\(\ge0\) chúng ta có đpcm.

Is that true?

WLOG \(b=mid\left\{a,b,c\right\}\). Áp dụng một bổ đề trong một bài giải của alibaba nguyễn trong câu hỏi của Neet ở học 24. Mọi người có thể tự chứng minh để nhớ lâu hoặc ai cần có thể hỏi ổng

\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{a+b}{b+c}+\frac{b+c}{a+b}+1\) với a,b,c>0

Khi đó ta cần chứng minh \(2\left(\frac{a+b}{b+c}+\frac{b+c}{a+b}\right)+2\ge\frac{2a+b+c}{b+c}+\frac{2b+c+a}{c+a}+\frac{2c+a+b}{a+b}\)

\(\Leftrightarrow\frac{a+b}{b+c}+\frac{b+c}{a+b}-\frac{1}{2}\ge\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(\Leftrightarrow\frac{b}{b+c}+\frac{b}{a+b}-\frac{1}{2}\ge\frac{b}{c+a}\)

\(\Leftrightarrow\frac{\left(a-b\right)\left(b-c\right)\left(a+c+2b\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)*đúng với \(b=mid\left\{a,b,c\right\}\)*

Lục vui câu hỏi của cô Chi thấy vài bài ngon mà mấy God dùng đao to vãi :))

\(\frac{1+a}{1-a}=\frac{1-a+2a}{1-a}=1+\frac{2a}{1-a}=1+\frac{2a}{b+c}\)

Khi đó BĐT cần chứng minh tương đương với:

\(3+\frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}\le2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)

\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3}{2}\le\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)

\(\Leftrightarrow\frac{ac}{b\left(b+c\right)}+\frac{bc}{a\left(a+b\right)}+\frac{ab}{c\left(c+a\right)}\ge\frac{3}{2}\)

Mặt khác:

\(LHS=\Sigma\frac{ac}{b\left(b+c\right)}=\Sigma\frac{a^2c^2}{abc\left(b+c\right)}\ge\frac{\left(ab+bc+ca\right)^2}{\Sigma abc\left(b+c\right)}=\frac{\left(ab+bc+ca\right)^2}{2abc\left(a+b+c\right)}\)

Ta cần chứng minh \(\frac{\left(ab+bc+ca\right)^2}{2abc\left(a+b+c\right)}\ge\frac{3}{2}\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\)

Tuy nhiên đây là bổ đề quen thuộc

Vậy ta có đpcm

dự đoán của Thần thánh

\(\frac{ab}{a^2+b^2}=\frac{a^2}{2a^2}=\frac{1}{2}\)

\(VT=\frac{3}{2}+\frac{9}{4}=\frac{12}{8}+\frac{18}{8}=\frac{30}{8}=\frac{15}{4}\)

\(p=\frac{ab}{a^2+b^2}+....+\frac{ca}{c^2+a^2};A=\frac{1}{4}\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\right)\)

áp dụng BDT cô si ta có

\(\frac{ab}{a^2+b^2}+\frac{\left(a^2+b^2\right)}{\frac{4}{9}}\ge2\sqrt{\frac{ab}{\frac{4}{9}}}=\frac{2}{\frac{2}{3}}\sqrt{ab}=3\sqrt{ab}\)

tương tự với các BDT còn lại suy ra

\(p+\frac{9}{4}\left(2a^2+2b^2+2c^2\right)\ge3\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\)

\(P+\frac{9}{2}\left(a^2+b^2+c^2\right)\ge3\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)

áp dụng BDT cô si ta có

\(a^2+\frac{1}{9}\ge2\sqrt{\frac{a^2}{9}}=\frac{2a}{3}\)

tương tự với b^2+c^2 ta được

\(a^2+b^2+c^2+\frac{1}{3}\ge\frac{2}{3}\left(a+b+c\right)=\frac{2}{3}\)

\(\Rightarrow a^2+b^2+c^2\ge\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\) 

" thay 1/3 vào ta được

\(p+\frac{3}{2}\ge3\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)

áp dụng BDT cô si dạng " Rei " " luôn đúng với những bài ngược dấu "

\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge3\sqrt[3]{\sqrt{abc}}=3\sqrt[3]{abc}\)

mà \(a+b+c\ge3\sqrt[3]{abc}\) 

thay a+b+c=1 vào ta được

\(P+\frac{3}{2}\ge3\Leftrightarrow P\ge\frac{6}{2}-\frac{3}{2}=\frac{3}{2}\) " 1 "

bây giờ tính nốt con \(A=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

áp dụng BDT \(\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\frac{1}{a+b+c}\)

\(A=\frac{9}{4}.\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\frac{9}{4}\left(\frac{1}{a+b+c}\right)\)

mà a+b+C=1 suy ra

\(A\ge\frac{9}{4}\) "2"

từ 1 và 2 suy ra

\(VT=P+A\ge\frac{3}{2}+\frac{9}{4}=\frac{12}{8}+\frac{18}{8}=\frac{30}{8}=\frac{15}{4}\)

" đúng với dự đoán của thần thánh "

\(S=\frac{2015-\left(a+b\right)}{a+b}+\frac{2015-\left(b+c\right)}{b+c}+\frac{2015-\left(a+c\right)}{a+c}=\frac{2015}{a+b}-\frac{a+b}{a+b}+\frac{2015}{b+c}-\frac{b+c}{b+c}+\frac{2015}{a+c}-\frac{a+c}{a+c}\)

\(S=2015.\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)-3=2015.\frac{1}{10}-3=\frac{1085}{10}\)

Ta có :

\(A+3=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+3\)

\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)

\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)

\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)

\(=2017.\frac{1}{2017}=1\)

\(\Rightarrow A=1-3=-2\)