1. a) \(a,b,c>0\). Cmr: \(\Sigma\frac{19b^3-a^3}{ab+5b^2}\le3\left(a+b+c\right)\)
b) \(\left\{{}\begin{matrix}a,b>0\\a^2+b^2\ge6\end{matrix}\right.\) . Cmr: \(\sqrt{3\left(a^2+6\right)}\ge\sqrt{2}\left(a+b\right)\)
2. a) \(\left\{{}\begin{matrix}x,y,z>0\\x^2\ge y^2+z^2\end{matrix}\right.\). Tìm Min \(A=x^2\left(\frac{1}{y^2}+\frac{1}{z^2}\right)+\frac{y^2+z^2}{x^2}\)
b) \(\left\{{}\begin{matrix}0\le a,b,c\le2\\a+b+c=3\end{matrix}\right.\). Tìm Max \(P=a^2+b^2+c^2\)
Ai bt giúp mk vs ! Mk cần trước 3h chiều nay ,Cảm ơn!
2a) Có cách này nhưng ko chắc!
\(A\ge\frac{4x^2}{y^2+z^2}+\frac{y^2+z^2}{x^2}=\frac{3x^2}{y^2+z^2}+\left(\frac{x^2}{y^2+z^2}+\frac{y^2+z^2}{x^2}\right)\)
\(\ge\frac{3\left(y^2+z^2\right)}{y^2+z^2}+2\sqrt{\frac{x^2}{y^2+z^2}.\frac{y^2+z^2}{x^2}}=3+2=5\)
Đẳng thức xảy ra khi x2 = y2 + z2????
tth, ?Amanda?, @Nk>↑@, buithianhtho, Phạm Hoàng Lê Nguyên,
Akai Haruma, Aki Tsuki, @Nguyễn Việt Lâm, @Trần Thanh Phương
Giúp mk vs!
1.
a) Ta thấy:
\(a^3+b^3-ab(a+b)=(a-b)^2(a+b)\geq 0, \forall a,b>0\)
\(\Rightarrow a^3+b^3\geq ab(a+b)\)
\(\Rightarrow \frac{19b^3-a^3}{ab+5b^2}=\frac{20b^3-(a^3+b^3)}{ab+5b^2}\leq \frac{20b^3-ab(a+b)}{ab+5b^2}=\frac{20b^2-a(a+b)}{a+5b}=\frac{(4b-a)(a+5b)}{a+5b}=4b-a\)
Hoàn toàn tương tự:
\(\frac{19c^3-b^3}{bc+5c^2}\leq 4c-b\); \(\frac{19a^3-c^3}{ac+5a^2}\leq 4a-c\)
Cộng theo vế và rút gọn:
\(\Rightarrow \sum \frac{19b^3-a^3}{ab+5b^2}\leq 3(a+b+c)\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
b) BĐT sai với $a=3,b=4$
2.
a) Áp dụng BĐT Cauchy-Schwarz:
\(A\geq x^2.\frac{4}{y^2+z^2}+\frac{y^2+z^2}{x^2}=\frac{3x^2}{y^2+z^2}+(\frac{x^2}{y^2+z^2}+\frac{y^2+z^2}{x^2})\)
Ta thấy:
\(\frac{3x^2}{y^2+z^2}\geq \frac{3(y^2+z^2)}{y^2+z^2}=3\); \(\frac{x^2}{y^2+z^2}+\frac{y^2+z^2}{x^2}\geq 2\) (AM-GM)
\(\Rightarrow A\geq 3+2=5\)
Vậy $A_{\min}=5$. Dấu "=" xảy ra khi \(\left\{\begin{matrix} y=z\\ x^2=y^2+z^2\end{matrix}\right.\Leftrightarrow x=\sqrt{2}y=\sqrt{2}z\)
Câu 2b:
\(P=a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=9-2(ab+bc+ac)\)
Vì \(a,b,c\leq 2\Rightarrow (a-2)(b-2)(c-2)\leq 0\)
\(\Leftrightarrow abc-2(ab+bc+ac)+4(a+b+c)-8\leq 0\)
\(\Rightarrow 2(ab+bc+ac)\geq abc+4(a+b+c)-8\geq 4(a+b+c)-8=4\) (do $abc\geq 0$ với mọi $a,b,c\geq 0$)
Do đó:
\(P=9-2(ab+bc+ac)\leq 9-4=5\)
Vậy $P_{\max}=5$ khi $(a,b,c)=(2,1,0)$ và hoán vị.
1b, Ta có: \(\left(a+b\right)^2=\left(\sqrt{2}.a\cdot\frac{1}{\sqrt{2}}+b.1\right)^2\le\left(2a^2+b^2\right)\left(\frac{1}{2}+1\right)\) (theo bđt bunhicopxki)
\(\Leftrightarrow\left(a+b\right)^2\le\left(a^2+6\right)\cdot\frac{3}{2}\) (do a^2+b^2=6)
\(\Leftrightarrow3\left(a^2+6\right)\ge2\left(a+b\right)^2\Leftrightarrow\sqrt{3\left(a^2+6\right)}\ge\left(a+b\right)\sqrt{2}\) (đpcm)
Dấu "=" ....
