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

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

\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)

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

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

\(\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)

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 $a=b=c$

Bài 2:

Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:

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

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

\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)

Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

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

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

\(VP\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=2\) (2)

(1);(2) \(\Rightarrow VT< VP\)

đb bị thiếu nhá bn, mik bổ sung ns sẽ thành: thỏa mãn a\(\le b\le c\)

Ta có \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\le\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{b}\)

bn tự chuyển vế quy đồng, sau đó ghép cặp nha

\(\dfrac{a^2\left(c-b\right)+b^2\left(a-c\right)+c^2\left(b-a\right)}{abc}\)

\(\dfrac{a^2\left(c-b\right)-b^2\left(a-c\right)+c^2\left(b-a\right)}{abc}\)

\(\dfrac{a^2\left(c-b\right)-b^2\left(c-b+b-a\right)+c^2\left(b-a\right)}{abc}\le0\)

\(\dfrac{a^2\left(c-b\right)-b^2\left(c-b\right)-b^2\left(b-a\right)+c^2\left(b-a\right)}{abc}0\le\)

\(\dfrac{\left(a-c\right)\left(a+b\right)\left(c-b\right)-\left(b-c\right)\left(b+c\right)\left(b-a\right)}{abc}\le0\)

\(\dfrac{\left(a-b\right)\left(c-b\right)\left[\left(a+b\right)-\left(b+c\right)\right]}{abc}\le0\)

\(\dfrac{\left(a-b\right)\left(c-b\right)\left(a-c\right)}{abc}\le0\)

Vì: \(0< a\le b\le c\) nên a-b <0; \(c-b\ge0\) \(a-c\le0\)

=>(a-b)(c-b)(a-c) \(\le\) 0 =>\(\dfrac{\left(a-b\right)\left(c-b\right)\left(a-c\right)}{abc}\le0\) ( đpcm)

Tích mình nhá, các bạn CTV hoặc thầy cô có thể kiểm tra lại xem em có làm đúng ko nhá ( đánh máy vội nên sẽ bị sai vài chỗ nên bn nhớ để ý nha )

Tham khảo nhé !