Triangle Inequality: (Bangladesh Online Math Camp)

[TOWFIQUL]

Anybody solved the exercise 1.10?

[Labib]

I was explaining your third case. So $y \geq 0$

[iPavel]

আমি মোবাইল থেকে ব্যাবহার করি তাই latex ব্যাবহার করতে পারিনা।

(উল্লেখ্য আপনার latex ব্যাবহার করে যে ইকুয়েশন গুলো লেখেন সেগুলো আমার মোবাইলে বিদঘুটে কিছু চিহ্ন আকারে দেখা যায়।এর কোন প্রতিকার আছে?)

[iPavel]

@toweikul
assume without loss of genarility,x greater than or equal y then (x)^1/2 is greater than or equal (y)^1/2

now consider two set,A={x,y} and B={(x)^1/2,(y)^1/2}

using rearrangement inequality,x(x)^1/2+y(y)^1/2 is greater than or equal x(y)^1/2+y(x)^1/2=(x)^1/2*(y)^1/2((x)^1/2+(y)^1/2)

[sampad]

This inequality is taught in class eleven in real numbers. But no teacher ever told us that it is called The Triangle inequality!

[*Mahi*]

It is called the triangle inequality because it can also be represented as the form of the classic geometric inequality ,
for $\triangle ABC$, sum of any two sides is greater than the third.

[mugdho.snigdho]

Could u plz state the problem 1.10 (i'm in the class & apart from the book right now ) ?

[nafistiham]

unfortunately, i couldn't solve this problem $1.10$ .again stating the problem for mugdho.snigdho and myself.hoping to get some hint

if $x,y> 0$, then prove that $\sqrt{\frac{x^{2}}{y}} + \sqrt{\frac{y^{2}}{x}}\geqslant \sqrt{x}+\sqrt{y}$

[Labib]

Tiham,
Would you like some hints?
Here are some if you want...

(a) Try squaring the sides and then simplify...
It should come to this-
It is enough to proof that,
$x^3+y^3\geq xy(x+y)$

(b) Now, without loss of generality, assume $x\geq y$,

(c) Now, substituting $x$ with $y+k$ follows the result.
The equality holds when $k=0$.

[Hasib]

To solve the proble 1.10, you have to define two case while $x \geq y$ and $x<y$ and u can try by squaring two side.