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Nyatmax

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Những câu trả lời của Nyatmax:

Vào lúc: 2020-02-06 18:11:23 Xem câu hỏi

\(\Sigma_{cyc}\frac{a^2-ab+bc+ca}{ab\left(a+b\right)\left(c+a\right)}\left(a-b\right)^2=\Sigma_{cyc}\frac{a-b+c+\frac{bc}{a}}{b\left(a+b\right)\left(c+a\right)}\left(a-b\right)^2\)

Theo tiêu chẩn 2 SOS thỉ \(VT-VP\ge0\)(chắc vậy)

Vào lúc: 2020-02-06 08:15:56 Xem câu hỏi

Bài ở dưới mình nhầm nhe.

Update

Ta có:

\(a^3+b^3+c^3\ge\frac{\left(a^2+b^2+c^2\right)^2}{a+b+c}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{3}\ge\frac{\left(a^2+b^2+c^2\right)\frac{\left(a+b+c\right)^2}{3}}{3}=a^2+b^2+c^2\)

\(\Rightarrow P\ge a^2+b^2+c^2+ab+bc+ca=\frac{a^2+b^2+c^2}{2}+\frac{\left(a+b+c\right)^2}{2}\ge\frac{\frac{\left(a+b+c\right)^2}{3}}{2}+\frac{9}{2}=6\)

Vào lúc: 2020-02-05 12:56:17 Xem câu hỏi

Ta có:

\(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)

\(\Rightarrow VT-VP=a^3+b^3+c^3+ab+bc+ca-6\ge a^3+b^3+c^3-ab-bc-ca\) (Giải thích:\(-6\ge-2\left(ab+bc+ca\right)\Rightarrow a^3+b^3+c^3+ab+bc+ca-6\ge a^3+b^3+c^3-ab-bc-ca\))

Ta lại có:

\(a^3+b^3+c^3-ab-bc-ca\ge\frac{\left(a^2+b^2+c^2\right)^2}{a+b+c}-\frac{\left(a+b+c\right)^2}{3}\ge\frac{\left[\frac{\left(a+b+c\right)^2}{3}\right]^2}{3}-3=0\)

\(\Rightarrow VT-VP\ge0\)

\(\Rightarrow P\ge6\)

Nếu có không đúng thì nhớ nói nhe chớ đừng có k sai tui giống mấy lần trước nhe :(

Vào lúc: 2020-02-01 19:59:14 Xem câu hỏi

WLOG \(c=min\left\{a;b;c\right\}\)

Let \(\hept{\begin{cases}a=x+c\\b=y+c\end{cases}\Rightarrow\hept{\begin{cases}x=a-c\\y=b-c\end{cases}}}\left(x,y\ge0\right)\)

Can chung minh

\(\frac{y+2c}{x+c}+\frac{x+2c}{y+c}+\frac{x+y+2c}{c}\ge4\left(\frac{x+c}{y+2c}+\frac{y+c}{x+2c}+\frac{c}{x+y+2c}\right)\)

\(VT-VP=8c^4\left(x^2-xy+y^2\right)+c^3\left(4x^3+10x^2y+10xy^2+4y^3\right)+c^2\left(6x^3y+24x^2y^2+6xy^3\right)+c\left(x^3y^2+x^2y^3\right)+2x^3y^3+x^2y^4\ge0\)

Vào lúc: 2020-01-28 12:04:05 Xem câu hỏi

\(A=\left(x+y\right)+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{1}{x}+\frac{1}{y}\right)+2\ge x+y+\frac{4}{x+y}+4\)

\(\Rightarrow A\ge\left(x+y+\frac{2}{x+y}\right)+\frac{2}{x+y}+4\ge2\sqrt{2}+4+\frac{2}{\sqrt{2\left(x^2+y^2\right)}}=3\sqrt{2}+4\)

Vào lúc: 2020-01-26 08:21:48 Xem câu hỏi

Cho o dong 2 la x,y,z nhe,ghi nham

Vào lúc: 2020-01-25 22:23:29 Xem câu hỏi

2.

Vỉ \(ab+bc+ca+abc=4\)thi luon ton tai \(a=\frac{2x}{y+z};b=\frac{2y}{z+x};c=\frac{2z}{x+y}\)

\(\Rightarrow VT=2\Sigma_{cyc}\sqrt{\frac{ab}{\left(b+c\right)\left(c+a\right)}}\le2\Sigma_{cyc}\frac{\frac{b}{b+c}+\frac{a}{c+a}}{2}=3\)

Vào lúc: 2020-01-19 08:03:50 Xem câu hỏi

67/3 nham

Vào lúc: 2020-01-19 08:03:16 Xem câu hỏi

\(x+1\ge3\Rightarrow\frac{67}{x+1}\le3\Rightarrow-\frac{67}{x+1}\ge3\)

Vào lúc: 2020-01-18 20:14:54 Xem câu hỏi

Repair

\(y=\left(2x^2+8\right)+\frac{5}{x+1}-8\ge8x+\frac{5}{x+1}-8\)

\(8x+\frac{5}{x+1}-8=8\left(x+1\right)+\frac{72}{x+1}-\frac{67}{x+1}-16\ge48-\frac{67}{3}-16=\frac{29}{3}\)

equelity iff \(x=2\)

Vào lúc: 2020-01-18 13:06:47 Xem câu hỏi

\(VT-VP=\Sigma_{cyc}\frac{a^2-ab+bc+ca}{ab\left(a+b\right)\left(c+a\right)}\left(a-b\right)^2\ge0\)

Equelity iff \(a=b=c\)

Vào lúc: 2020-01-18 12:43:03 Xem câu hỏi

\(y=\left(2x^2+\frac{16}{x}+\frac{16}{x}\right)-\frac{27}{x}+1\ge24-\frac{27}{2}+1=\frac{23}{2}\)

Equelity iff \(x=2\)

Vào lúc: 2020-01-16 20:36:39 Xem câu hỏi

\(VT\ge\frac{9}{\Sigma_{cyc}\sqrt{xy+x+y}}\ge\frac{9}{\sqrt{\left(1+1+1\right)\left(2x+2y+2z+xy+yz+zx\right)}}\ge\frac{9}{\sqrt{3\left[6+\frac{\left(x+y+z\right)^2}{3}\right]}}=\sqrt{3}\)

Vào lúc: 2020-01-15 13:48:37 Xem câu hỏi

Repair đề \(\Sigma_{cyc}\frac{\sqrt{2a^2+b^2}}{ab}\ge3\sqrt{3}\).Because dấu '=' xảy ra khi \(a=b=c=3\)

Không use condition của đề bài :))

Ta co:

\(VT=\sqrt{\frac{a}{b}+\frac{a}{b}+\frac{b}{a}}+\sqrt{\frac{b}{c}+\frac{b}{c}+\frac{c}{b}}+\sqrt{\frac{c}{a}+\frac{c}{a}+\frac{a}{c}}\)

\(\Rightarrow VT\ge\sqrt{3\sqrt[3]{\frac{a}{b}}}+\sqrt{3\sqrt[3]{\frac{b}{c}}}+\sqrt{3\sqrt[3]{\frac{c}{a}}}\ge3\sqrt[3]{\sqrt{3\sqrt[3]{\frac{a}{b}}.\sqrt{3\sqrt[3]{\frac{b}{c}}.\sqrt{3\sqrt[3]{\frac{c}{a}}}}}}=3\sqrt{3}\)

equelity iff \(a=b=c=3\)

Vào lúc: 2020-01-09 20:55:43 Xem câu hỏi

We have:

\(A=\Sigma_{cyc}\frac{1}{3xy+3zx+x+y+z}\le\frac{1}{3xy+3zx+3\sqrt[3]{xyz}}=\Sigma_{cyc}\frac{1}{3xy+3zx+3}=\Sigma_{cyc}\frac{1}{3\left(xy+zx+1\right)}\)

Dat \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\Rightarrow abc=1\)

\(\Rightarrow A\le\Sigma_{cyc}\frac{1}{3\left(\frac{1}{ab}+\frac{1}{ca}+1\right)}=\Sigma_{cyc}\frac{a}{3\left(a+b+c\right)}=\frac{1}{3}\)

Dau '=' xay ra khi \(x=y=z=1\)

Vào lúc: 2020-01-06 13:17:48 Xem câu hỏi

Sau mot hoi mo

\(f\left(a;b;c\right)=a^3+b^3+c^3+3abc-ab\left(a+b\right)-bc\left(b+c\right)-ca\left(c+a\right)\)

\(=\frac{1}{2}\left(b+c-a\right)\left(a-b\right)^2+\frac{1}{2}\left(c+a-b\right)\left(b-c\right)^2+\frac{1}{2}\left(a+b-c\right)\left(c-a\right)^2\)

Wlog \(a\ge b\ge c\)

\(\hept{\begin{cases}c+a-b\ge0\\\left(b+c-a\right)+\left(c+a-b\right)\ge0\\\left(c+a-b\right)+\left(a+b-c\right)\ge0\end{cases}}\)

Vào lúc: 2020-01-05 14:19:57 Xem câu hỏi

\(\text{Condition}:x,y\ge0\)

\(\hept{\begin{cases}x^2+2x=4-\sqrt{y}\left(M_1\right)\\y^2+2y=4-\sqrt{x}\left(M_2\right)\end{cases}}\)

\(\left(M_1\right)-\left(M_2\right)\Leftrightarrow\left(x^2-y^2\right)+2\left(x-y\right)+\left(\sqrt{x}-\sqrt{y}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)+2\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)+\left(\sqrt{x}-\sqrt{y}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y\\\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)+2\left(\sqrt{x}+\sqrt{y}\right)+1=0\left(M_3\right)\end{cases}}\)

x=0 khong phai nghiem PT\(\Rightarrow M_3\)(fail)

Thay x=y vao 

:D

Vào lúc: 2020-01-04 20:53:21 Xem câu hỏi

Dat \(\left(\frac{a}{b};\frac{b}{c};\frac{c}{a}\right)=\left(x;y;z\right)\)

\(\Rightarrow xyz=1\)

\(\Sigma_{cyc}\frac{1}{\frac{a}{b}+\frac{c}{a}+1}=\Sigma_{cyc}\frac{1}{x+y+1}\)

We need to prove:

\(\Sigma_{cyc}\frac{1}{x+y+1}\le1\)

\(\Leftrightarrow\Sigma_{cyc}\frac{x+y}{x+y+1}\ge2\left(M\right)\)

We have:

\(VT_M\ge\frac{\left(\Sigma_{cyc}\sqrt{x+y}\right)^2}{2\Sigma_{cyc}x+3}\)

Now we need to prove

\(\frac{\left(\Sigma_{cyc}\sqrt{x+y}\right)^2}{2\Sigma_{cyc}x+3}\ge2\)

\(\Leftrightarrow\Sigma_{cyc}\sqrt{\left(x+y\right)\left(y+z\right)}\ge\Sigma_{cyc}x+3\left(M_1\right)\)

Consider:

\(VT_{M_1}=\sqrt{\left(x+y\right)\left(y+z\right)}\ge x+y+z+xy+yz+zx\)

Now we need to prove:

\(x+y+z+xy+yz+zx\ge x+y+z+3\)

\(xy+yz+zx\ge3\) (Not fail with xyz=1)

Dau '=' xay ra khi \(\hept{\begin{cases}a=b=c=1\\x=y=z=1\end{cases}}\)

Vào lúc: 2020-01-04 20:41:06 Xem câu hỏi

Repair de

\(\frac{\left(a+b\right)^2}{2c^2+a^2+b^2}+\frac{\left(b+c\right)^2}{2a^2+b^2+c^2}+\frac{\left(c+a\right)^2}{2c^2+a^2+b^2}\)

\(\Rightarrow VT=\Sigma_{cyc}\frac{\left(a+b\right)^2}{\left(c^2+a^2\right)+\left(b^2+c^2\right)}\le\Sigma_{cyc}\left(\frac{a^2}{c^2+a^2}+\frac{b^2}{b^2+c^2}\right)=3\)

Dau '=' xay ra khi \(a=b=c=1\)

Vào lúc: 2020-01-04 20:21:38 Xem câu hỏi

\(\text{Condition}:x>y\)

HPT\(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)\left(x^2+y^2\right)=20\\\left(x+y\right)^2\left(x-y\right)=32\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x^2+y^2\right)\left(x-y\right)=20\\\frac{5}{8}\left(x+y\right)^2\left(x-y\right)=20\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x^2+y^2\right)\left(x-y\right)=20\left(1\right)\\\left(x-y\right)\left[\left(x^2+y^2\right)-\frac{5}{8}\left(x+y\right)^2\right]=0\left(2\right)\end{cases}}\)

(2)\(\Leftrightarrow\orbr{\begin{cases}x=y\\\frac{3}{8}x^2+\frac{3}{8}y^2-\frac{5}{4}xy=0\left(3\right)\end{cases}}\)

y=0 khong phai nghiem cua HPT

(3)\(\Leftrightarrow\frac{x^2}{y^2}-\frac{5x}{4y}+\frac{3}{8}=0\)

\(\Rightarrow\Delta=\frac{1}{16}>0\left(t=\frac{x}{y}>1\right)\)

\(\Rightarrow2x=3y;x=y\)

Thay vao roi tinh :D

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