a. Xét hiệu: \(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}\)

=\(\dfrac{b\left(a+b\right)+a\left(a+b\right)-4ab}{ab\left(a+b\right)}\)

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

Vì a,b>0

Xảy ra đẳng thức khi và chỉ khi a=b

a) Ta có: \(\left(a-b\right)^2\ge0\left(1\right)\forall a,b\)

( Dấu = xày ra khi và chỉ khi a=b)

Cộng 4ab vào 2 vế, ta có:

\(\left(a-b\right)^2+4ab\ge4ab\Leftrightarrow\left(a+b\right)^2\ge4ab\)

Chia 2 vế cho ab(a+b)>0, ta có:

\(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\Leftrightarrow\)\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

b) Ta có:

\(2p=a+b+c\)

\(p-a=\dfrac{a+b+c}{2}-a=\dfrac{b+c-a}{2}>0\) vì b+c>a

Tương tự: \(p-b>0,p-c>0\)

Áp dụng BĐT: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)cho từng cặp số p-a, p-b; p-b,p-c;p-c,p-a

Ta có:

\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{\left(p-a\right)+\left(p-b\right)}=\dfrac{4}{2p-\left(a+b\right)}=\dfrac{4}{c}\left(1\right)\)

Tương tự:

\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{a}\left(2\right)\)

\(\dfrac{1}{p-c}+\dfrac{1}{p-a}\ge\dfrac{4}{b}\left(3\right)\)

Cộng các BĐT cùng chiều (1), (2), (3) vế theo vế, ta có:

\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Do đó: \(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

BĐT cô si: \(\dfrac{x+y}{2}>\left(hoặc=\right)\sqrt{xy}\)

=>x+y >(hoặc =) \(2\sqrt{xy}\)

=>\(\left(x+y\right)^2>\left(hoặc=\right)4xy\)

=>\(\dfrac{1}{x}+\dfrac{1}{y}>\left(hoặc=\right)\dfrac{4}{x+y}\)

vì P=\(\dfrac{a+b+c}{2}=>a+b+c=2p\)

=>c=2p-a-b

b=2p-a-c

a=2p-b-c

ta có:\(\dfrac{1}{p-a}+\dfrac{1}{p-b}>hoặc=\dfrac{4}{p-a+p-b}=\dfrac{4}{c}\)

\(\dfrac{1}{p-a}+\dfrac{1}{p-c}>\left(hoặc=\right)\dfrac{4}{p-a+p-c}=\dfrac{4}{b}\)

\(\dfrac{1}{p-b}+\dfrac{1}{p-c}>\left(hoặc=\right)\dfrac{4}{p-b+p-c}=\dfrac{4}{a}\)

cộng vế với vế ta đc

\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)>\left(hoặc=\right)4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

<=>\(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}>\left(hoặc=\right)2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Dễ dàng chứng minh bất đẳng thức phụ :

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\forall a;b>0\)và p - a; p - b; p - c > 0 theo bất đẳng thức trong tam giác.

Áp dụng bất đẳng thức phụ vừa chứng minh, ta có:

\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{2p-a-b}=\dfrac{4}{c}\left(1\right)\)

\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{2p-b-c}=\dfrac{4}{a}\left(2\right)\)

\(\dfrac{1}{p-c}+\dfrac{1}{p-a}\ge\dfrac{4}{2p-c-a}=\dfrac{4}{a}\left(3\right)\)

Cộng (1); (2); (3) theo vế, ta có:

\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\RightarrowĐPCM\)

Ta CM BĐT sau :

\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Thật vậy ; ta có :

\(\left(x-y\right)^2\ge0\\ \Rightarrow x^2-2xy+y^2\ge0\\ \Rightarrow x^2+y^2\ge2xy\\ \Rightarrow\left(x+y\right)^2\ge4xy\\ \Rightarrow\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\\ \Rightarrow\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\left(đpcm\right)\)

\(\Rightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{2p-\left(a+b\right)}=\dfrac{4}{c}\\ \dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{a}\\ \dfrac{1}{p-a}+\dfrac{1}{p-c}\ge\dfrac{4}{b}\\ \Rightarrow2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\\ \Rightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(đpcm\right)\)

Áp dụng bđt Cauchy-Schwarz:

\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{\left(1+1\right)^2}{2p-a-b}=\dfrac{4}{c}\)

\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{\left(1+1\right)^2}{2p-b-c}=\dfrac{4}{a}\)

\(\dfrac{1}{p-a}+\dfrac{1}{p-c}\ge\dfrac{\left(1+1\right)^2}{2p-a-c}=\dfrac{4}{b}\)

Cộng theo vế:

\(2VT\ge4VP\Leftrightarrow VT\ge2VP\Leftrightarrowđpcm\)

\("="\Leftrightarrow a=b=c\)

a, Áp dụng BĐT Cosi:

\(\sqrt{\left(p-a\right)\left(p-b\right)}\le\dfrac{p-a+p-b}{2}=\dfrac{c}{2}\)

\(\sqrt{\left(p-b\right)\left(p-c\right)}\le\dfrac{p-b+p-c}{2}=\dfrac{a}{2}\)

\(\sqrt{\left(p-c\right)\left(p-a\right)}\le\dfrac{p-c+p-a}{2}=\dfrac{b}{2}\)

\(\Rightarrow\left(p-a\right)\left(p-b\right)\left(p-c\right)\le\dfrac{1}{8}abc\)

b, \(\dfrac{r}{R}=\dfrac{\dfrac{S_{ABC}}{p}}{\dfrac{abc}{4S_{ABC}}}\)

\(=\dfrac{4S_{ABC}^2}{p.abc}=\dfrac{4.p\left(p-a\right)\left(p-b\right)\left(p-c\right)}{p.abc}\)

\(\le\dfrac{4.p.\dfrac{1}{8}abc}{p.abc}=\dfrac{1}{2}\)

c, Áp dụng BĐT Cosi:

\(a.m_a=\dfrac{2\sqrt{3}}{3}.\dfrac{\sqrt{3}}{2}a.m_a\)

\(\le\dfrac{2\sqrt{3}}{3}.\dfrac{\dfrac{3}{4}a^2+m_a^2}{2}\)

\(=\dfrac{\sqrt{3}}{3}.\left(\dfrac{3}{4}a^2+\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}\right)\)

\(=\dfrac{\sqrt{3}\left(a^2+b^2+c^2\right)}{6}\)

\(\Rightarrow a.m_a\le\dfrac{\sqrt{3}\left(a^2+b^2+c^2\right)}{6};b.m_b\le\dfrac{\sqrt{3}\left(a^2+b^2+c^2\right)}{6};c.m_c\le\dfrac{\sqrt{3}\left(a^2+b^2+c^2\right)}{6}\)

Khi đó \(\dfrac{a}{m_a}+\dfrac{b}{m_b}+\dfrac{c}{m_c}\)

\(=\dfrac{a^2}{a.m_a}+\dfrac{b^2}{b.m_b}+\dfrac{c^2}{c.m_c}\)

\(\ge\dfrac{a^2+b^2+c^2}{\dfrac{\sqrt{3}\left(a^2+b^2+c^2\right)}{6}}=2\sqrt{3}\)

Câu hỏi của Phạm Thị Hường - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo bài làm ở link này nhé!