The second direction is not always true, in a BST all elements in the right subtree of a node should have a key bigger than its key and all elements in the left subtree of a node should have a key less than its key. for example, consider this binary tree:

4 / \ 2 6 / \ / \ 1 9 3 7

it has the property that for every node x of $T$ that is not a leaf, the key of x is larger or equal than the key of the left child and less or equal than the key of the right child, but it's not a BST since 3 is in the right subtree of root but its key is less than the key of root.
(also you used greater or equal in your sentence, but in BST, keys should be identical, so if two keys are equal it can't be a BST anyway)

The second direction is not always true, in a BST all elements in the right subtree of a node should have a key bigger than its key and all elements in the left subtree of a node should have a key less than its key. for example, consider this binary tree:

4 / \ 2 6 / \ / \ 1 9 3 7

it has the property that for every node x of $T$ that is not a leaf, the key of x is larger or equal than the key of the left child and less or equal than the key of the right child, but it's not a BST since 3 is in the right subtree of root but its key is less than the key of root.