a) 4a^2b^2 > (a^2+b^2-c^2)^2 với a,b,c là độ dài 3 cạnh của một tam giác.
b) a^4+b^4+2 ≥ 4ab (a,b>0)
c) (a+b)(a^3+b^3)≤ 2(a^4+b^4) (a,b>0)
d) bc/a+ca/b+ab/c ≥ a+b+c (a,b,c>= 0)
e) ab/(a+b) +bc/(b+c)+ca/(c+a) ≤ (a+b+c)/2 (a,b,c>0)
Tks nha!!
[tex]a)a+b>c\\\Leftrightarrow (a+b)^2>c^2\\\Leftrightarrow a^2+b^2+2ab>c^2\\\Leftrightarrow 2ab>c^2-(a^2+b^2)\\\Leftrightarrow (2ab)^2>[c^2-(a^2+b^2)]^2\\\Leftrightarrow 4a^2b^2>(a^2+b^2-c^2)^2[/tex]
$b)$Áp dụng BĐT Cô-si cho 2 số dương
[tex]\Rightarrow a^4+b^4\geq 2\sqrt{a^4.b^4}=2a^2b^2\\\Rightarrow a^4+b^4+2\geq 2a^2b^2+2\geq 2\sqrt{2a^2b^2.2}=2.2ab=4ab[/tex]
[tex]c)(a+b)(a^3+b^3)\leq 2(a^4+b^4)\\\Leftrightarrow a^4+b^4+a^3b+ab^3\leq 2(a^4+b^4)\\\Leftrightarrow a^3b+ab^3\leq a^4+b^4\\\Leftrightarrow a^4-a^3b+b^4-ab^3\geq 0\\\Leftrightarrow a^3(a-b)-b^3(a-b)\geq 0\\\Leftrightarrow (a-b)(a^3-b^3)\geq 0\\\Leftrightarrow (a-b)^2(a^2+ab+b^2)\geq 0(luôn \ đúng)[/tex]
$d)$Áp dụng BĐT Cô-si cho 2 số dương
[tex]\Rightarrow \dfrac{bc}{a}+\dfrac{ca}{b}\geq 2\sqrt{\dfrac{bc}{a}.\dfrac{ac}{b}}=2c\\Tương \ tự:\dfrac{ca}{b}+\dfrac{ab}{c}\geq 2a;\dfrac{ab}{c}+\dfrac{bc}{a}\geq 2b\\\Rightarrow 2\left ( \dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c} \right )\geq 2(a+b+c)\\\Leftrightarrow \dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}\geq a+b+c[/tex]
[tex]e)(a+b)^2\geq 4ab\Leftrightarrow \dfrac{a+b}{4}\geq \dfrac{ab}{a+b}\\(b+c)^2\geq 4bc\Leftrightarrow \dfrac{b+c}{4}\geq \dfrac{bc}{b+c}\\(c+a)^2\geq 4ca\Leftrightarrow \dfrac{c+a}{4}\geq \dfrac{ca}{c+a}\\\Rightarrow \dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\leq \dfrac{a+b}{4}+\dfrac{b+c}{4}+\dfrac{c+a}{4}=\dfrac{2(a+b+c)}{4}=\dfrac{a+b+c}{2}[/tex]
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