Theo bất đẳng thức tam giác
\(\Rightarrow\left\{\begin{matrix}a< b+c\\b< c+a\\c< a+b\end{matrix}\right.\Rightarrow\left\{\begin{matrix}b+c-a>0\\c+a-b>0\\a+b-c>0\end{matrix}\right.\)
Áp dụng bất đẳng thức \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\forall a,b>0\)
\(\Rightarrow\left\{\begin{matrix}\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{2}{b}\\\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}\ge\dfrac{2}{c}\\\dfrac{1}{a+b-c}+\dfrac{1}{a+c-b}\ge\dfrac{2}{a}\end{matrix}\right.\)
Cộng theo từng vế
\(\Rightarrow2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ( đpcm )
Câu 1: mik sửa đề tí
Ta có: a+b=1
a² +b² ≥ (a+b)²/2
<=> a² +b² ≥ 1/2(a² +b²) + ab
<=> 1/2(a² +b²) -ab ≥ 0
<=> 1/2(a-b)² ≥ 0 ( luôn đúng )
vậy a² + b² ≥ (a+b)²/2 = 1/2
tương tự thì
a^4 + b^4 ≥ (a² +b²)²/2 ≥ (1/2)²/2 = 1/8
vậy a^4 + b^4 ≥ 1/8
dấu = xảy ra <=> a=b=1/2