\(\dfrac{a}{a+b+c}>\dfrac{a}{a+b+c+d}\)

Làm tương tự với 3 phân số còn lại và cộng vế với vế

\(\dfrac{a}{a+b+c}< \dfrac{a+d}{a+b+c+d}\)

Làm tương tự với 3 phân số còn lại và cộng vế với vế

\(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)

\(\Leftrightarrow a\left(b+c\right)< b\left(a+c\right)\)

\(\Leftrightarrow ab+ac< ba+bc\)

\(\Leftrightarrow ac< bc\)

\(\Leftrightarrow a< b\)(đúng)

a)Áp dụng

\(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}=2\left(1\right)\)

Lại có:\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a}{a+b+c}+\dfrac{b}{b+c+a}+\dfrac{c}{c+a+b}=1\left(2\right)\)

Từ (1) và (2)=> đpcm

Vì \(\dfrac{a}{b}< 1\Rightarrow a< b\Rightarrow ac< bc\Rightarrow ac+ab< bc+ab\Rightarrow a\left(b+c\right)< b\left(a+c\right)\Rightarrow\dfrac{a\left(b+c\right)}{b\left(b+c\right)}< \dfrac{b\left(a+c\right)}{b\left(b+c\right)}\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+c}\)a) ta có

\(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}\)\(\Leftrightarrow\dfrac{a+b+c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{2\left(a+b+c\right)}{a+b+c}\)

\(\Leftrightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)

a: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)

\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}\)

Do đó: \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)

b: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)

DO đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

Bài 1: Nhân chéo

Bài 2:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)

\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}\)

\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)

\(\Rightarrowđpcm\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}\)

\(=\dfrac{a+b+c-a+b-c}{a+b-c-a+b+c}\)

\(=\dfrac{\left(a-a\right)+\left(b+b\right)+\left(c-c\right)}{\left(a-a\right)+\left(b+b\right)+\left(c-c\right)}\)

\(=\dfrac{2b}{2b}=1\)

\(\Rightarrow a+b+c=a+b-c\)

\(\Rightarrow c=-c\)

\(\Rightarrow c+c=0\)

\(\Rightarrow2c=0\Rightarrow c=0\)

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(1\right)\)

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3\)

\(=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:

\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)

Bài 1:

\(P=(x+1)\left(1+\frac{1}{y}\right)+(y+1)\left(1+\frac{1}{x}\right)\)

\(=2+x+y+\frac{x}{y}+\frac{y}{x}+\frac{1}{x}+\frac{1}{y}\)

Áp dụng BĐT Cô-si:

\(\frac{x}{y}+\frac{y}{x}\geq 2\)

\(x+\frac{1}{2x}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)

\(y+\frac{1}{2y}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)

Áp dụng BĐT SVac-xơ kết hợp với Cô-si:

\(\frac{1}{2x}+\frac{1}{2y}\geq \frac{4}{2x+2y}=\frac{2}{x+y}\geq \frac{2}{\sqrt{2(x^2+y^2)}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)

Cộng các BĐT trên :

\(\Rightarrow P\geq 2+2+\sqrt{2}+\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)

Vậy \(P_{\min}=4+3\sqrt{2}\Leftrightarrow a=b=\frac{1}{\sqrt{2}}\)

Bài 2:

Áp dụng BĐT Svac-xơ:

\(\frac{1}{a+3b}+\frac{1}{b+a+2c}\geq \frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)

\(\frac{1}{b+3c}+\frac{1}{b+c+2a}\geq \frac{4}{2b+4c+2a}=\frac{2}{b+2c+a}\)

\(\frac{1}{c+3a}+\frac{1}{c+a+2b}\geq \frac{4}{2c+4a+2b}=\frac{2}{c+2a+b}\)

Cộng theo vế và rút gọn :

\(\Rightarrow \frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\geq \frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\) (đpcm)

Dấu bằng xảy ra khi $a=b=c$

Tham khảo:Chứng minh a/b=c/d hoặc a/b=d/c biết (a^2+b^2)/(c^2+d^2)=ab/cd - An Nhiên

\(\text{Cho }\dfrac{a}{b}=\dfrac{d}{c}\text{ và }b,d\notin0\text{.CMR:}\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

\(\text{Ta có:}\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\text{Lại có:}\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{\left(bd\right).k^2}{bd}=k^2\)

\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{\left(b^2+d^2\right).k^2}{b^2+d^2}=k^2\)

\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

Ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

a) \(\dfrac{3a+5c}{3b+5d}=\dfrac{3\cdot bk+5\cdot dk}{3b+5d}=\dfrac{k\left(3b+5d\right)}{3b+5d}=k\) (1)

\(\dfrac{a-2c}{b-2d}=\dfrac{bk-2dk}{b-2d}=\dfrac{k\left(b-2d\right)}{b-2d}=k\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{3a+5c}{3b+5d}=\dfrac{a-2c}{b-2d}\left(dpcm\right)\)

b) \(\dfrac{a^2-b^2}{ab}=\dfrac{\left(bk\right)^2-b^2}{bk\cdot b}=\dfrac{b^2k^2-b^2}{b^2k}=\dfrac{b^2\left(k-1\right)}{b^2k}=\dfrac{k-1}{k}\)(1)

\(\dfrac{c^2-d^2}{cd}=\dfrac{\left(dk\right)^2-d^2}{dk\cdot d}=\dfrac{d^2k^2-d^2}{d^2k}=\dfrac{d^2\left(k-1\right)}{d^2k}=\dfrac{k-1}{k}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\left(dpcm\right)\)

c) \(\left(\dfrac{a+b}{c+d}\right)^3=\left(\dfrac{bk+b}{dk+d}\right)^3=\dfrac{b^3\left(k+1\right)^3}{d^3\left(k+1\right)^3}=\dfrac{b^3}{d^3}\) (1)

\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\dfrac{b^3}{d^3}\) (2)

Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^3=\dfrac{a^3+b^3}{c^3+d^3}\left(dpcm\right)\)

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