a)a<b

=>a+c<b+c(1)

c<d

=>b+c<b+d(2)

Từ 1 và 2 =>a+c<b+d

b)a<b

=>ac<bc(1)

c<d

=>bc<bd(2)

Từ 1 và 2 =>ac<bd

a) a<b \(\Rightarrow\) a+c < b+c (1)

c<d\(\Rightarrow\) c+b < d+b (2)

Từ 1 và 2 \(\Rightarrow\)a+c < b+d (dpcm)

b) a<b \(\Rightarrow\) ac < bc ( vì c dương) (1)

c < d\(\Rightarrow\) bc < bd (vì b dương) (2)

Từ 1 và 2 \(\Rightarrow\) ac < bd (đpcm)

a)Áp dụng AM-GM có:

\(a\sqrt{b-1}\le a.\dfrac{b-1+1}{2}=\dfrac{ab}{2}\)

\(b\sqrt{a-1}\le b.\dfrac{a-1+1}{2}=\dfrac{ab}{2}\)

\(\Rightarrow a\sqrt{b-1}+b\sqrt{a-1}\le\dfrac{ab}{2}+\dfrac{ab}{2}\)

\(\Leftrightarrow a\sqrt{b-1}+b\sqrt{a-1}\le ab\)

Dấu "=" xảy ra khi a=b=2

b)Áp dụng bđt bunhiacopxki có:

\(\left(\sqrt{ac}+\sqrt{bd}\right)^2=\left(\sqrt{a}.\sqrt{c}+\sqrt{b}.\sqrt{d}\right)^2\)\(\le\left[\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2\right]\left[\left(\sqrt{c}\right)^2+\left(\sqrt{d}\right)^2\right]=\left(a+b\right)\left(c+d\right)\)

\(\Rightarrow\sqrt{ac}+\sqrt{bd}\le\sqrt{\left(a+b\right)\left(c+d\right)}\)

Dấu "=" xảy ra khi \(\dfrac{\sqrt{a}}{\sqrt{c}}=\dfrac{\sqrt{b}}{\sqrt{d}}\Leftrightarrow ad=bc\)

\(b,\) Áp dụng BĐT Bunhiacopski:

\(\left(a+b\right)\left(c+d\right)=\left[\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2\right]\left[\left(\sqrt{c}\right)^2+\left(\sqrt{d}\right)^2\right]\\ \ge\left(\sqrt{ac}+\sqrt{bd}\right)^2\)

Dấu \("="\Leftrightarrow ad=bc\)

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

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

a: \(\Leftrightarrow\left(a+1\right)^2-4a\ge0\)

hay \(\left(a-1\right)^2>=0\)(luôn đúng)

b: \(VT=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)

\(=\left(c^2+d^2\right)\left(a^2+b^2\right)=VP\)

a: Xét tứ giác ABCD có 

\(\widehat{A}+\widehat{C}=180^0\)

Do đó: ABCD là tứ giác nội tiếp