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\(\dfrac{a.h_a}{2}=S\Leftrightarrow a=\dfrac{2S}{h_a}\)
Tương tự:
\(b=\dfrac{2S}{h_b};c=\dfrac{2S}{h_c}\)
\(\dfrac{a+b+c}{4S}=\dfrac{\dfrac{2S}{h_a}+\dfrac{2S}{h_b}+\dfrac{2S}{h_c}}{4S}=\dfrac{2S\left(\dfrac{1}{h_a}+\dfrac{1}{h_b}+\dfrac{1}{h_c}\right)}{4S}=\dfrac{\dfrac{1}{h_a}+\dfrac{1}{h_b}+\dfrac{1}{h_c}}{2}\)
Tương đương:
\(\dfrac{1}{h_a+h_b}+\dfrac{1}{h_b+h_c}+\dfrac{1}{h_c+h_a}\le\dfrac{\dfrac{1}{h_a}+\dfrac{1}{h_b}+\dfrac{1}{h_c}}{2}\)
Cauchy-Schwarz:
\(\dfrac{1}{h_a+h_b}\le\dfrac{1}{4}\left(\dfrac{1}{h_a}+\dfrac{1}{h_b}\right)\)
\(\dfrac{1}{h_b+h_c}\le\dfrac{1}{4}\left(\dfrac{1}{h_b}+\dfrac{1}{h_c}\right)\)
\(\dfrac{1}{h_c+h_a}\le\dfrac{1}{4}\left(\dfrac{1}{h_c}+\dfrac{1}{h_a}\right)\)
Cộng theo vế suy ra đpcm
Theo đề bài thì ta có:
\(ah_a=bh_b=ch_c=2\)
Ta có:
\(\left(a^2+b^2+c^2\right)\left(h_a^2+h_b^2+h_c^2\right)\ge\left(ah_a+bh_b+ch_c\right)^2\)
\(=\left(2+2+2\right)^2=36\)
Dấu = xảy ra khi \(\hept{\begin{cases}a=b=c=\frac{2}{\sqrt[4]{3}}\\h_a=h_b=h_c=\sqrt[4]{3}\end{cases}}\)
A B C a b c
Có \(\sin\widehat{A}=\frac{h_c}{b}=\frac{h_b}{c}=\frac{h_c-h_b}{b-c}=\frac{h_b-h_c}{\frac{a}{k}}=\frac{k\left(h_b-h_c\right)}{a}\) (1)
Lại có : \(\hept{\begin{cases}\sin\widehat{B}=\frac{h_c}{a}\\\sin\widehat{C}=\frac{h_b}{a}\end{cases}}\)\(\Rightarrow\)\(k\left(\sin\widehat{B}-\sin\widehat{C}\right)=\frac{k\left(h_c-h_b\right)}{a}\) (2)
(1) (2) ...
\(\sin\widehat{B}=\frac{h_a}{c}\)\(;\)\(\sin\widehat{C}=\frac{h_a}{b}\) (1)
\(\hept{\begin{cases}\sin\widehat{B}=\frac{h_c}{a}\\\sin\widehat{C}=\frac{h_b}{a}\end{cases}\Leftrightarrow\hept{\begin{cases}h_c=\sin\widehat{B}.a\\h_b=\sin\widehat{C}.a\end{cases}}}\)\(\Rightarrow\)\(k\left(\frac{1}{h_b}-\frac{1}{h_c}\right)=\frac{k}{a}.\left(\frac{1}{\sin\widehat{C}}-\frac{1}{\sin\widehat{B}}\right)\) (2)
Thay (1) vào (2) ta được \(\frac{k}{a}.\left(\frac{1}{\sin\widehat{C}}-\frac{1}{\sin\widehat{B}}\right)=\frac{k}{a}.\left(\frac{b}{h_a}-\frac{c}{h_a}\right)=\frac{k}{a}.\frac{\frac{a}{k}}{h_a}=\frac{1}{h_a}\)
đpcm