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Bài 4:
Ta có:Vì a,b,c là độ dài 3 cạnh của 1 tam giác nên a+b-c>0,a+c-b>0,b+c-a>0.Do đó,áp dụng bất thức \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)với x,y là các số dương
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{a+b-c}+\frac{1}{a+c-b}\ge\frac{4}{\left(a+b-c\right)+\left(a+c-b\right)}=\frac{4}{2a}=\frac{2}{a}\\\frac{1}{a+b-c+}+\frac{1}{b+c-a}\ge\frac{4}{\left(a+b-c\right)+\left(b+c-a\right)}=\frac{4}{2b}=\frac{2}{b}\\\frac{1}{b+c-a}+\frac{1}{a+c-b}\ge\frac{4}{\left(b+c-a\right)+\left(a+c-b\right)}=\frac{4}{2c}=\frac{2}{c}\end{matrix}\right.\)
\(\Rightarrow2\left(\frac{1}{b+c-a}+\frac{1}{a+c-b}+\frac{1}{a+b-c}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Mà \(\left\{{}\begin{matrix}b+c-a=\left(a+b+c\right)-2a=2p-2a=2\left(p-a\right)\\a+c-b=\left(a+b+c\right)-2b=2p-2b=2\left(p-b\right)\\a+b-c=\left(a+b+c\right)-2c=2p-2c=2\left(p-c\right)\end{matrix}\right.\)
\(\Rightarrow2\left[\left(\frac{1}{2\left(p-a\right)}+\frac{1}{2\left(p-b\right)}+\frac{1}{2\left(p-c\right)}\right)\right]\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(đpcm\right)\)
Dấu "=" xảy ra khi và chỉ khi a=b=c
5.
\(\sqrt{\frac{x}{y+z}}=\frac{x}{\sqrt{x\left(y+z\right)}}\ge\frac{2x}{x+y+z}\)
Tương tự: \(\sqrt{\frac{y}{x+z}}\ge\frac{2y}{x+y+z}\) ; \(\sqrt{\frac{z}{x+y}}\ge\frac{2z}{x+y+z}\)
Cộng vế với vế:
\(VT\ge\frac{2\left(x+y+z\right)}{x+y+z}=2\)
Dấu "=" ko xảy ra nên \(VT>2\)
\(B=\frac{ab}{a+b+2}\Rightarrow2B=\frac{2ab}{a+b+2}=\frac{\left(a+b\right)^2-a^2-b^2}{a+b+2}=\frac{\left(a+b\right)^2-4}{a+b+2}=a+b-2\)
Do a ; b không âm , áp dụng BĐT Cô - si cho 2 số , ta có :
\(a+b\le\sqrt{2\left(a^2+b^2\right)}=\sqrt{2.4}=\sqrt{8}\)
\(\Rightarrow a+b-2\le\sqrt{8}-2\)
\(\Rightarrow2B\le\sqrt{8}-2\Rightarrow B\le\sqrt{2}-1\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=\sqrt{2}\)
Do x ; y không âm , \(x^2+y^2=1\)
\(\Rightarrow\left|x\right|;\left|y\right|\le1\) \(\Rightarrow0\le x;y\le1\)
\(\Rightarrow x\ge x^2;y\ge y^2\Rightarrow x+y\ge x^2+y^2=1\)
\(x,y\ge0\Rightarrow xy\ge0\)
Ta có : \(A=\sqrt{5x+4}+\sqrt{5y+4}\)
\(\Rightarrow A^2=5x+4+5y+4+2\sqrt{\left(5x+4\right)\left(5y+4\right)}\)
\(=5\left(x+y\right)+8+2\sqrt{25xy+20y+20x+16}\)
\(\ge5.1+8+2\sqrt{25.0+20.1+16}=13+2.6=25\)
\(\Rightarrow A\ge5\)
Dấu " = " xảy ra \(\Leftrightarrow\left[{}\begin{matrix}x=0;y=1\\x=1;y=0\end{matrix}\right.\)
1/ Đầu tiên ta chứng minh: \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\) (1)
\(\Leftrightarrow\Sigma_{cyc}\left(\frac{a^3}{b}-a^2\right)\ge0\Leftrightarrow\Sigma_{cyc}\left(\frac{a^2\left(a-b\right)}{b}-a\left(a-b\right)\right)+\Sigma_{cyc}a\left(a-b\right)\ge0\)
\(\Leftrightarrow\Sigma_{cyc}\frac{a\left(a-b\right)^2}{b}+\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\)
\(\Leftrightarrow\Sigma_{cyc}\frac{a\left(a-b\right)^2}{b}+\Sigma_{cyc}\frac{1}{2}\left(a-b\right)^2\ge0\)
\(\Leftrightarrow\Sigma_{cyc}\frac{\left(a-b\right)^2\left(2a+b\right)}{2b}\ge0\)
BĐT cuối đúng nên (1) đúng. (*)
Bây giờ ta đi chứng minh: \(a^2+b^2+c^2\ge5\)
Đặt \(\left(a+b+c;ab+bc+ca\right)\rightarrow\left(3u;3v^2\right)\) thì \(3u=9-3v^2\)
và \(a^2+b^2+c^2=\left(3u\right)^2-6v^2=\left(9-3v^2\right)^2-6v^2\)
\(=\left(3v^2-9\right)^2-6v^2=9v^4-60v^2+81\)
Đặt \(v^2=t\ge0\) .Ta cần tìm min của: \(9t^2-60t+81\)
Ta có: \(9t^2-60t+81=\left(3t-10\right)^2-19\ge-19\)
Dấu "=" xảy ra khi t = 10/3 tức là v= \(\sqrt{\frac{10}{3}}\)....
Em thấy có gì đó sai sai thì phải ạ:((
Câu 1:
\(\frac{a^3}{b}+ab\ge2a^2\) ; \(\frac{b^3}{c}+bc\ge2b^2\); \(\frac{c^3}{a}+ac\ge2c^2\)
\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+ab+ac+bc\ge2\left(a^2+b^2+c^2\right)\)
\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-\left(ab+ac+bc\right)\)
\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-\left(a^2+b^2+c^2\right)=a^2+b^2+c^2\)
//
\(a+b+c+ab+ac+bc\le a+b+c+\frac{\left(a+b+c\right)^2}{3}\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{3}+\left(a+b+c\right)\ge9\)
\(\Rightarrow\left(a+b+c-\frac{3\sqrt{13}-3}{2}\right)\left(a+b+c+\frac{3\sqrt{13}+3}{2}\right)\ge0\)
\(\Rightarrow a+b+c\ge\frac{3\sqrt{13}-3}{2}\)
\(\Rightarrow a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\ge\frac{1}{3}\left(\frac{3\sqrt{13}-3}{2}\right)^2=\frac{21-3\sqrt{13}}{2}>5\)
\(\Rightarrow a^2+b^2+c^2>5\)
Dấu "=" ko xảy ra
Ta có \(3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)
Do đó \(Q\ge\frac{\left(x+y+z\right)^2}{3}+\frac{1}{x+y+z}=\frac{\left(x+y+z\right)^2}{3}+\frac{9}{x+y+z}+\frac{9}{x+y+z}-\frac{17}{x+y+z}\)
\(\ge3\sqrt[3]{\frac{9\cdot9\cdot\left(x+y+z\right)^2}{3\cdot\left(x+y+z\right)^2}}-\frac{17}{x+y+z}\ge9-\frac{17}{3\sqrt[3]{xyz}}=9-\frac{17}{3}=\frac{10}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Lời giải:
Áp dụng BĐT AM-GM:
$\sqrt{x}+\sqrt{x}+x^2\geq 3\sqrt[3]{x^3}=3x$
$\sqrt{y}+\sqrt{y}+y^2\geq 3y$
$\sqrt{z}+\sqrt{z}+z^2\geq 3z$
Cộng theo vế:
$2(\sqrt{x}+\sqrt{y}+\sqrt{z})+x^2+y^2+z^2\geq 3(x+y+z)=(x+y+z)^2$
$\Leftrightarrow 2(\sqrt{x}+\sqrt{y}+\sqrt{z})\geq 2(xy+yz+xz)$
$\Leftrightarrow \sqrt{x}+\sqrt{y}+\sqrt{z}\geq xy+yz+xz$
Ta có đpcm.
Dấu "=" xảy ra khi $x=y=z=1$
ĐK:y\(\ge0\)
\(P=x^2-x\sqrt{y}+x+y-\sqrt{y}+1=\left(x^2-x\sqrt{y}+\dfrac{y}{4}+x-\dfrac{\sqrt{y}}{2}+\dfrac{1}{4}\right)+\left(\dfrac{3}{4}y-\dfrac{\sqrt{y}}{2}+\dfrac{1}{12}\right)+\dfrac{2}{3}=\left(x-\dfrac{\sqrt{y}}{2}+\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}y-\dfrac{\sqrt{3}}{6}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\forall x\in R;y\ge0\)
=>Min P=\(\dfrac{2}{3}\)đạt được khi \(\left\{{}\begin{matrix}x-\dfrac{\sqrt{y}}{2}+\dfrac{1}{2}=0\\\dfrac{\sqrt{3}}{2}y-\dfrac{\sqrt{3}}{6}=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=\dfrac{\sqrt{3}-3}{6}\\y=\dfrac{1}{3}\end{matrix}\right.\)
Úi lộn làm lại nha
ĐK:y\(\ge0\)
\(P=x^2-x\sqrt{y}+x+y-\sqrt{y}+1=\left(x^2-x\sqrt{y}+\dfrac{y}{4}-\dfrac{\sqrt{y}}{2}+\dfrac{1}{4}+x\right)+\left(\dfrac{3}{4}y-\dfrac{\sqrt{y}}{2}+\dfrac{1}{12}\right)+\dfrac{2}{3}=\left(x-\dfrac{\sqrt{y}}{2}+\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3y}}{2}-\dfrac{\sqrt{3}}{6}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\forall x\in R;y\ge0\)
=>Min P=\(\dfrac{2}{3}\)đạt được khi \(\left\{{}\begin{matrix}x-\dfrac{\sqrt{y}}{2}+\dfrac{1}{2}=0\\\dfrac{\sqrt{3y}}{2}-\dfrac{\sqrt{3}}{6}=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=\dfrac{1}{9}\end{matrix}\right.\)
\(\hept{\begin{cases}\left(x+y\right)^2=16\\\left(x-y\right)^2\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2+2xy+y^2=16\\x^2-2xy+y^2\ge0\end{cases}\Leftrightarrow}x^2+y^2\ge8}\)
áp dụng AM - GM có:
\(P=x^2+y^2+\frac{12}{xy}\ge x^2+y^2+\frac{12}{\frac{x^2+y^2}{2}}=8+\frac{2.12}{8}=14\)
Vậy \(P_{min}\)=14 dấu "=" sảy ra khi : x=x=2
Bài giải sai rồi