1.

 Áp dụng BĐT BSC:

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

Đẳng thức xảy ra khi \(a=b=c>0\)

2.

Áp dụng BĐT \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\) và BĐT BSC:

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

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

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

\(\le2.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)\)

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

Đẳng thức xảy ra khi \(a=b=c>0\)

Cách khác:

1.

 Áp dụng BĐT Cauchy:

\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}+\dfrac{b^2}{c+a}+\dfrac{c+a}{4}+\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\)

Đẳng thức xảy ra khi \(a=b=c>0\)

Câu 1:

a, Giả sử \(A=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-\dfrac{a}{b}-\dfrac{b}{a}\ge0\)

\(\Leftrightarrow A=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-2\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge0\)

Mà \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\Leftrightarrow A\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-2\cdot\dfrac{a}{b}-2\cdot\dfrac{b}{a}+2\ge0\)

\(\Leftrightarrow\left(\dfrac{a^2}{b^2}-2\cdot\dfrac{a}{b}+1\right)+\left(\dfrac{b^2}{a^2}-2\cdot\dfrac{b}{a}+1\right)\ge0\\ \Leftrightarrow\left(\dfrac{a}{b}-1\right)^2+\left(\dfrac{b}{a}-1\right)^2\ge0\left(\text{luôn đúng}\right)\)

Dấu \("="\Leftrightarrow a=b\)

b, \(B=\dfrac{a^4}{b^4}+\dfrac{b^4}{a^4}-2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\right)+2+\left(\dfrac{a^2}{b^2}+2+\dfrac{b^2}{a^2}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)-4\)

\(B=\left(\dfrac{a^4}{b^4}-2\cdot\dfrac{a^2}{b^2}+1\right)+\left(\dfrac{b^4}{a^4}-2\cdot\dfrac{b^2}{a^2}+1\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)-2\\ \Leftrightarrow B=\left(\dfrac{a^2}{b^2}-1\right)^2+\left(\dfrac{b^2}{a^2}-1\right)^2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2+\dfrac{a}{b}+\dfrac{b}{a}-4\\ \Leftrightarrow B\ge0+0+0+\dfrac{a^2+b^2}{ab}-4\ge\dfrac{2ab}{ab}-4=2-4=-2\)

Dấu \("="\Leftrightarrow\left(a;b\right)\in\left\{\left(1;-1\right);\left(-1;1\right)\right\}\)

Câu 2:

\(\left(x^2+y^2\right)\left(3^2+4^2\right)\ge\left(3x+4y\right)^2=M^2\\ \Leftrightarrow M^2\le25\cdot25\\ \Leftrightarrow M\le25\)

Dấu \("="\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{25}{25}=1\Leftrightarrow\left\{{}\begin{matrix}x^2=9\\y^2=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

Vậy \(M_{max}=25\Leftrightarrow\left(x;y\right)=\left(3;4\right)\)

Áp dụng BĐT Cô si dạng phân số ta có :

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

=> ĐPCM .

b) Vì a,b,c > 0 .

Áp dụng BĐT Cô si ta có :

\(\dfrac{a^2}{b}+b\ge2a\) (1)

Tương tự ta có : \(\dfrac{b^2}{c}+c\ge2b\) (2)

\(\dfrac{c^2}{a}+a\ge2c\) (3)

Cộng từng vế => ĐPCM .

a)\(\dfrac{a}{b}=5-\dfrac{3}{5}=\dfrac{25}{5}-\dfrac{3}{5}=\dfrac{22}{5}\)

b)\(\dfrac{a}{b}=\dfrac{5}{6}+\dfrac{4}{7}=\dfrac{35}{42}+\dfrac{24}{42}=\dfrac{59}{42}\)

c)\(\dfrac{a}{b}=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{3}{5}\times\dfrac{3}{2}=\dfrac{9}{10}\)

d)\(\dfrac{a}{b}=3\times\dfrac{2}{7}=\dfrac{6}{7}\)

e)\(\dfrac{a}{b}=\dfrac{7}{5}-\left(\dfrac{2}{5}\times\dfrac{1}{2}\right)=\dfrac{7}{5}-\dfrac{1}{5}=\dfrac{6}{5}\)

a 22/5

b 59/42

c 9/10

d 6/7

e 6/5

a.

Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)

Tương tự: \(\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge b\) ; \(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)

Cộng vế:

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

\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

b.

Ta có:

\(a^2+bc\ge2\sqrt{a^2bc}=2\sqrt{ab.ac}\Rightarrow\dfrac{1}{a^2+bc}\le\dfrac{1}{2\sqrt{ab.ac}}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{ac}\right)\)

Tương tự: \(\dfrac{1}{b^2+ac}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{bc}\right)\) ; \(\dfrac{1}{c^2+ab}\le\dfrac{1}{4}\left(\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)

Cộng vế với vế:

\(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{1}{2}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{a+b+c}{2abc}\)

Dấu "=" xảy ra khi \(a=b=c\)

a,Từ \(\dfrac{a}{c}=\dfrac{c}{b}\)⇒\(c^2=a.b\)

Khi đó \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a^2+a.b}{b^2+a.b}\\ =\dfrac{a\left(a+b\right)}{b\left(a+b\right)}\)

b,Ta có:

\(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\Rightarrow\dfrac{b^2+c^2}{a^2+c^2}=\dfrac{a}{b}\\ \dfrac{a^2+c^2}{b^2+c^2}=\dfrac{b}{a}\Rightarrow\dfrac{b^2+c^2}{a^2+c^2}-1=\dfrac{b}{a}-1\\ hay\dfrac{b^2+c^2-a^2-c^2}{a^2+c^2}=\dfrac{b-a}{a}\)

Vậy \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)

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 $a=b=c=\frac{1}{3}$