a) xét ΔABC ta có

C<A

=> AB < BC ( quan hệ giữa góc và cạnh đối diện trong Δ)

b)xét ΔABD ta có

BD = BA

=> ΔABD là Δ cân tại B

mà B=60^o

=> ΔABD làΔ đều

Ta chứng minh:\(\sqrt{a+bc}\ge a+\sqrt{bc}\)

\(\Leftrightarrow a+bc\ge a^2+bc+2a\sqrt{bc}\)

\(\Leftrightarrow a\ge a^2+2a\sqrt{bc}\)\(\Leftrightarrow a\ge a\left(a+2\sqrt{bc}\right)\Leftrightarrow1\ge a+2\sqrt{bc}\Leftrightarrow a+b+c\ge a+2\sqrt{bc}\)

\(\Leftrightarrow b+c-2\sqrt{bc}\ge0\Leftrightarrow\left(\sqrt{b}-\sqrt{c}\right)^2\ge0\)(luôn đúng)

\(\Leftrightarrow\sqrt{a+bc}\ge a+\sqrt{bc}\)

CMTT\(\sqrt{b+ca}\ge b+\sqrt{ca}\)

          \(\sqrt{c+ab}\ge c+\sqrt{ab}\)

\(\Leftrightarrow\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=1+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)Vậy ......

(Dấu = xảy ra (=) a=b=c=1/3

\(\frac{a-bc}{a+bc}=\frac{a-bc}{a\left(a+b+c\right)+bc}=\frac{a-bc}{a^2+ab+bc+ca}=\frac{a-bc}{\left(a+b\right)\left(c+a\right)}\)

\(=\left(a-bc\right)\sqrt{\frac{1}{\left(a+b\right)^2\left(c+a\right)^2}}\le\frac{\frac{a-bc}{\left(a+b\right)^2}+\frac{a-bc}{\left(c+a\right)^2}}{2}=\frac{a-bc}{2\left(a+b\right)^2}+\frac{a-bc}{2\left(c+a\right)^2}\)

Tương tự, ta có: \(\frac{b-ca}{b+ca}\le\frac{b-ca}{2\left(b+c\right)^2}+\frac{b-ca}{2\left(a+b\right)^2}\)\(;\)\(\frac{c-ab}{c+ab}\le\frac{c-ab}{2\left(c+a\right)^2}+\frac{c-ab}{2\left(b+c\right)^2}\)

=> \(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\le\frac{a-bc+b-ca}{2\left(a+b\right)^2}+\frac{b-ca+c-ab}{2\left(b+c\right)^2}+\frac{a-bc+c-ab}{2\left(c+a\right)^2}\)

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

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

hay a=b=c

ad t/ch dãy tỉ số bằng nhau, ta có:

a b = b c = c a = a + b + c b + c + a = 1 ⇒ a = b = c ( dpcm )

Từ điều kiện đề bài ta có  a b + b c + c a a b c = 3 ⇔ 1 a + 1 b + 1 c = 3  

Áp dụng hai lần bất đẳng thức Côsi cho hai số dương, ta có:

a 2 + b c ≥ 2 a 2 . b c = 2 a b c ⇒ a a 2 + b c ≤ 2 2 a b c = 1 2 b c 1 b . 1 c ≤ 1 2 1 b + 1 c ⇒ a a 2 + b c ≤ 1 4 1 b + 1 c

Tương tự ta có: 

b b 2 + c a ≤ 1 4 1 c + 1 a ; c c 2 + a b ≤ 1 4 1 a + 1 b ⇒ a a 2 + b c + b b 2 + c a + c c 2 + a b ≤ 1 2 1 a + 1 b + 1 c = 3 2 .