By Cauchy-Schwarz's inequality:

\(L.H.S=\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\)

\(=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{bc+ab}+\dfrac{c^4}{ac+bc}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}\)

\(\ge\dfrac{a^2+b^2+c^2}{2}=\dfrac{1}{2}=R.H.S\)

DONE!

This pro will be solved by SOS method

\(\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}\ge\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(\Leftrightarrow\dfrac{a^2b+a^2c-ab^2-ac^2}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{ab^2+b^2c-a^2b-bc^2}{\left(a^2+c^2\right)\left(a+c\right)}+\dfrac{bc^2+ac^2-a^2c-b^2c}{\left(a^2+b^2\right)\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{ab\left(b-a\right)+bc\left(c-b\right)}{\left(a^2+c^2\right)\left(a+c\right)}+\dfrac{bc\left(c-b\right)+ac\left(c-a\right)}{\left(a^2+b^2\right)\left(a+b\right)}\ge0\)

\(\LeftrightarrowΣ\left(\dfrac{ab\left(a-b\right)}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{ab\left(b-a\right)}{\left(a^2+c^2\right)\left(a+c\right)}\right)\ge0\)

\(\LeftrightarrowΣ\left(ab\left(a-b\right)\left(\dfrac{\left(a^2+c^2\right)\left(a+c\right)-\left(b^2+c^2\right)\left(b+c\right)}{\left(b^2+c^2\right)\left(a^2+c^2\right)\left(b+c\right)\left(a+c\right)}\right)\right)\ge0\)

\(\LeftrightarrowΣ\left(ab\left(a-b\right)\cdot\dfrac{\left(a-b\right)\left(a^2+b^2+c^2+ab+bc+ca\right)}{\left(b^2+c^2\right)\left(a^2+c^2\right)\left(b+c\right)\left(a+c\right)}\right)\ge0\)

\(\LeftrightarrowΣ\left(ab\left(a-b\right)^2\dfrac{a^2+b^2+c^2+ab+bc+ca}{\left(b^2+c^2\right)\left(a^2+c^2\right)\left(b+c\right)\left(a+c\right)}\right)\ge0\) *RIGHT!*

I have just finshed it Here

Exercise 2:

a)\(4x-x^2-5\)

\(=-x^2+4x-4-1\)

\(=-\left(x^2-4x+4\right)-1\)

\(=-\left(x-2\right)^2-1\le-1\)

When \(x=2\)

b)\(5-x^2-6x\)

\(=-x^2-6x-9+14\)

\(=-\left(x^2+6x+9\right)+14\)

\(=-\left(x+3\right)^2+14\le14\)

When \(x=-3\)

Search GTNN, GTLN @@

a)\(x^2-4x+1\)

\(=\left(x^2-4x+4\right)-3\)

\(=\left(x-2\right)^2-3\ge-3\)

When \(x=2\)

b)\(3x^2-6x-1\)

\(=3x^2-6x+3-4\)

\(=3\left(x^2-2x+1\right)-4\)

\(=3\left(x-1\right)^2-4\ge-4\)

When \(x=1\)

c)\(4x^2+4x-2\)

\(=4x^2+4x+1-3\)

\(=\left(2x+1\right)^2-3\ge-3\)

When \(x=-\dfrac{1}{2}\)

Use Cauchy-Schwarz's inequality we have:

\(L.H.S=\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\)

\(=\dfrac{1}{\dfrac{a+b}{ab}}+\dfrac{1}{\dfrac{b+c}{bc}}+\dfrac{1}{\dfrac{c+a}{ca}}\)

\(=\dfrac{ab}{a+b}+\dfrac{ac}{a+c}+\dfrac{bc}{b+c}\)

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

\(\le a+b+c-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}\)

\(=a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}=R.H.S\)

When \(a=b=c\)

Cảm ơn  ! Các bạn đã gõ ra những gì mà cả Topic này biết

\(75+12+43+25+88+50+57\)

\(=\left(75+25\right)+\left(12+88\right)+\left(43+57\right)+50\)

\(=100+100+100+50\)

\(=300+50=350\)

Use AM-GM's ineq:

\(\dfrac{b+c}{\sqrt{a}}+\dfrac{c+a}{\sqrt{b}}+\dfrac{a+b}{\sqrt{c}}\ge\dfrac{2\sqrt{bc}}{\sqrt{a}}+\dfrac{2\sqrt{ca}}{\sqrt{b}}+\dfrac{2\sqrt{ab}}{\sqrt{c}}\)

\(=\left(\sqrt{\dfrac{bc}{a}}+\sqrt{\dfrac{ca}{b}}\right)+\left(\sqrt{\dfrac{ca}{b}}+\sqrt{\dfrac{ab}{c}}\right)+\left(\sqrt{\dfrac{ab}{c}}+\sqrt{\dfrac{bc}{a}}\right)\)

\(\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b}+\sqrt{c}\)

\(\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{abc}\)\(=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)

Example:  they're \(\angle B=\angle M;\angle C=\angle Q;\angle A=\angle P\) but they aren't same. This is Similar triangles

i have a solution but it's ugly

If \(a=b=1\) and \(c=0\)   then we get a value  \(2+\frac{1}{\sqrt2}\)

We'll prove that it's a minimal value. Thus, we need to prove that

\(\sqrt{\dfrac{ab+bc+ca}{a^2+b^2}}+\sqrt{\dfrac{ab+bc+ca}{b^2+c^2}}+\sqrt{\dfrac{ab+bc+ca}{a^2+c^2}}\ge2+\dfrac{1}{\sqrt{2}}\)

WLOG \(c=\min\{a,b,c\}\). Hence:

\(\dfrac{ab+ac+bc}{a^2+b^2}-\dfrac{(a+c)(b+c)}{(a+c)^2+(b+c)^2}=\dfrac{c(a+b+2c)(2ab+ac+bc)}{a^2+b^2)((a+c)^2+(b+c)^2}\ge0\)

Similar\(\dfrac{ab+ac+bc}{a^2+c^2}-\dfrac{b+c}{a+c}=\dfrac{c(2ab+ac-c^2)}{(a+c)(a^2+c^2)}\ge0\)

And \(\dfrac{ab+ac+bc}{b^2+c^2}-\dfrac{a+c}{b+c}=\dfrac{c(2ab+bc-c^2)}{(b+c)(b^2+c^2)}\ge0\)

Let \(\dfrac{a+c}{b+c}=x^2;\dfrac{b+c}{a+c}=y^2\left(x,y>0\right)\)\(\Rightarrow xy=1\) and we have:

\(x+y+\dfrac{1}{\sqrt{x^2+y^2}}\ge2+\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow x+y-2\sqrt{xy}\ge\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{x^2+y^2}}\)

\(\Leftrightarrow(\sqrt{x}-\sqrt{y})^2\ge\dfrac{(x-y)^2}{\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})}\)

\(\Leftrightarrow\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})\ge(\sqrt{x}+\sqrt{y})^2\)

By Cauchy-Schwarz's ine we have: 

\(\sqrt{2(x^2+y^2)}=\sqrt{(1^2+1^2)(x^2+y^2)}\ge x+y\)

\(=\dfrac{1}{2}(1^2+1^2)((\sqrt{x})^2+(\sqrt{y})^2)\ge\dfrac{1}{2}(\sqrt{x}+\sqrt{y})^2\)

Thus, it's enough to prove that \(\sqrt{x^2+y^2}+\sqrt{2}\ge2\)

It's true by AM-GM \(\sqrt{x^2+y^2}+\sqrt{2}\ge\sqrt{2xy}+\sqrt{2}=2\sqrt{2}>2\)

Use AM-GM's inequality we have:

\(\dfrac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{abc}{2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}}\)\(=\dfrac{abc}{8abc}=\dfrac{1}{8}\)

cube not square. this is a cube 

\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}\) wrong too

\(\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\) - it's wrong

Auto Ask Auto Answer ? i will report it :|

WLOG \(x\le y \le z\) we have:

\(\dfrac{7}{10}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{z}=\dfrac{3}{z}\)

\(\Rightarrow\dfrac{7}{10}\ge\dfrac{3}{z}\Rightarrow z\ge\dfrac{30}{7}\approx4,28\)

Because \(z\) is positive integer id est \(z\ge 5\)

Because \(x+y+z\) is smallest value should find \(z\) is smallest value

\(\Rightarrow z=5\Rightarrow x=3;y=6\) (wrong because \(x\le y \le z\))

\(\Rightarrow z=6\Rightarrow y=5\Rightarrow x=3\) (right because \(x\le y \le z\))

We only find 1 smallest value satisfy, should \(z=7;z=8;... \) wrong

Hence \(x+y+z=3+5+6=14\)

\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)\(\Leftrightarrow x+3=308\Rightarrow x=305\)

the equation don't have solution i think it's wrong

From \(3a^2+3b^2=10ab\)

\(\Rightarrow3a^2+3b^2-10ab=0\)

\(\Rightarrow3a^2-9ab-ab+3b^2=0\)

\(\Rightarrow3a\left(a-3b\right)-b\left(a-3b\right)=0\)

\(\Rightarrow\left(3a-b\right)\left(a-3b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3a-b=0\\a-3b=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}b=3a\\a=3b\end{matrix}\right.\)

*)\(b=3a\Rightarrow P=\dfrac{a-b}{a+b}=\dfrac{a-3a}{a+3a}=\dfrac{-2}{4}=-\dfrac{1}{2}\)

*)\(a=3b\Rightarrow P=\dfrac{a-b}{a+b}=\dfrac{3b-b}{3b+b}=\dfrac{2}{4}=\dfrac{1}{2}\)